Keluarga Kami Seneby

Rasanya baru kemarin kita duduk bersama di kelas memakai seragam putih abu.

Dengan berbagai kisah dan cerita suka-duka.

Sudah lebih dari 9 tahun kita bersama melewati berbagai rona kehidupan kita masing-masing.

Apalagi bila teringat insiden UN itu tuh? Hadeuh. Wkwkkwk.

Hampura dulur-dulur sadaya. T_T

Apabila aku rindu kalian, kupandangi lagi album kenangan SMA kita yang penuh warna itu.

Kok bisa ya kita dulu seperti itu? Heu.,heu.,heu.,

Kok bisa ya kita sampai saat ini bisa terus bersalam-silaturahmi?

Meski "pasukan" Seneby tak lengkap hadir semoga kekeluargaan ini tetap langgeng abadi.

Mendengarkan kisah-kisah para sahabat yang telah memiliki keberhasilan di ranahnya membuat hati ini sangat gembira dan bangga.

Maafkan segala khilaf dan salah ku sobat semua.

Selamat berkarya di "ladang" masing-masing Kawan.

Seneby We Are Family

Together Forever

Aku di sini hanya dapat mendoakan yang terbaik, semoga apa-apa yang kalian cita-citakan dan impikan dapat teraih serta mendapat Ridha-Nya.

Semoga hingga anak cucu kita kelak persahabatan dan kekeluargaan ini tetap Allah pelihara.


Aamiin.

Terima Kasih Kawan-Kawan ku.

Pendidikan Sebagai Elemen Pertahanan Bangsa

"Membangun sistem pembelajaran yang baik dan menyeluruh serta didampingi dengan konsep wawasan ke-bangsaan akan memperkokoh persatuan-kesatuan sebuah bangsa"

Setiap bangsa dalam mempertahankan eksistensi dan mewujudkan cita-citanya perlu memiliki pemahaman mengenai geopolitik dan geostrategi. Geopolitik bangsa Indonesia diterjemahkan dalam konsep Wawasan Nusantara, sedangkan geostrategi bangsa Indonesia dirumuskan dalam konsep Ketahanan Nasional. 

Sesuai dengan bagan paradigma ketatanegaraan Negara Republik Indonesia, maka Ketahanan Nasional (Tannas) merupakan salah satu konsepsi politik dari Negara Republik Indonesia. Ketahanan Nasional dapat dikatakan sebagai konsep geostrateginya bangsa Indonesia. 

Dengan kata lain, geostrategi bangsa Indonesia diwujudkan melalui konsep Ketahanan Nasional. Untuk meresepkan konsep-konsep ini perlu diadakan upaya edukasi masyarakat yang kreatif dengan berbagai metode, misalnya melalui pendidikan, perfilman dan acara-acara nasional yang bersifat mendidik masyarakat agar mencintai bangsa dan negaranya.

Semoga

Aamiin.

Sumber:

Arip Nurahman Notes

Misi Mendampingi SBMPTN Accomplished

Selama lebih dari seminggu ini penulis dan para sahabat belajar menjadi panitia Ujian SBMPTN  Tahun 2013.

Belajar menjadi pengawas Ujian SBMPTN 2013 bagian SOSHUM di SMAN 22 Bandung bersama sahabat karib kang Reza Ruhbani, S.Pd.

Sebelum hari H pelaksanaan Ujian SBMPTN kami di berikan pengarahan dan pelatihan terlebih dahulu oleh salah seorang Guru Besar (Profesor) Biologi dari UPI Bandung tentang bagaimana menjadi seorang pengawas yang baik, ketika beliau sedang menjelaskan tata cara menjadi seorang pengawas sungguh sangat lucu sekali, karena beliau sangat pintar dan cerdas dalam memaparkan strategi untuk mejadi pengawas yang jempolan.

"Awas hati-hati kepada para pengawas jangan main mata dengan peserta ujian SBMPTN" katanya.

"Waspada apabila melihat gerak-gerik para peserta Ujian yang mencurigakan, tahu kan? Kalian pasti tahu ya?" Sambungnya.

Hadeuhhhh.,. ada.,ada sajah.,dech pak Prof.

Wkwkwkkwkw.

Banyak sekali pengalaman yang didapat, diantaranya bertemu dengan bapa ibu guru yang sudah berpengalaman di SMAN 22 Bandung, bertukar wawasan serta pengalaman.

Berbincang dan bertukar pikiran dengan Koordinator Lokasi (KORLOK) Bpk. Dr. Irwan Meilano, ST, M.Sc., seorang ahli Gempa bumi (Ph.D.) lulusan dari Earth Science, Nagoya University, Japan.

Allhamdulilah beliau memberikan wejangan sekaligus motivasi untuk senantiasa belajar dan tentu saja diakhir tugas, beliau memberikan kartu nama serta Honor bagi kita semua yang cukup untuk membuat "dapur" kami "ngebul" selama sebulan.

Terima Kasih dan Allhamdulilah yah.

He.,he.he.,.

Bertemu juga dengan peserta ujian dari SMAN 1 dan 2 Banjar, saya berpesan kepada mereka untuk terus berjuang meraih cita-citanya, mereka ingin meneruskan Pendidikan ke Jurusan Ilmu Pemerintahan UNPAD, dan Jurusan Manajemen Pariwisata Universitas Pendidikan Indonesia, tak lupa di akhir Ujian saya titip salam bagi bapa ibu guru di SMAN 1 dan 2 Banjar.

Adik-adikku Semoga cita kalian Allah berikan jalan terbaik. 

KETERANGAN INFORMASI PANLOK BANDUNG

Panitia Lokal Bandung menginformasikan pelaksanaan Ujian Tertulis SBMPTN 2013 dan berbagai data terkait penerimaan mahasiswa baru tahun 2013 di empat Perguruan Tinggi Negeri (PTN) di Kota Bandung. 

Menurut Sekretaris Eksekutif SBMPTN Panlok Bandung, Prof. Asep Gana Suganda, Ph.D.  dari 40 ribu total rencana peserta yang mendaftar SBMPTN, sebanyak 37.682 peserta mengikuti SBMPTN di hari pertama. Dari jumlah tersebut yakni 14.709 peserta dari Kelompok Ujian Saintek, 15.809 peserta kelompok ujian SOSHUM dan sisanya 7.164 dari kelompok ujian campuran. 

Sementara total kuota penerimaan mahasiswa baru di empat PTN yang ada di Kota Bandung bervariasi jumlahnya. Prof. Asep merinci bahwa. Kuota SBMPTN ITB sejumlah 1.372, Unpad sejumlah 2.010, UPI sejumlah 2.084, dan UIN SGD sejumlah 1.536. "Unpad tidak membuka Ujian Mandiri," ujar Doktor Farmasi lulusan Universite de Nantes Perancis tersebut. 

Sementara peminat PTN terbesar dari SBMPTN Panlok Bandung adalah Unpad dengan 73.557 peserta. Sementara bertutut-turut UPI, ITB dan UIN dengan jumlah peminat masing-masing 48.784 peserta, 28.075 peserta dan 6.808 peserta.

Impian Pendidikan Tinggi di Indonesia

Saya bermimpi di masa depan seluruh pelajar yang ingin melanjutkan ke jenjang pendidikan setinggi apa pun dapat mereka lakukan tanpa terkendala biaya dan kekurangan fasilitas apa pun.

Semoga.

Ucapan Terima Kasih:

Prof. Ahmaloka, Ph.D.
Ketua Umum SBMPTN 2013 dan Rektor ITB.

Prof. H. Furqon, M.Si., M.Sc., Ph.D.
Ketua PANLOK SBMPTN 2013 Bandung, Dosen Metodologi Penelitian Pendidikan UPI, Bandung.

Teh Siti Latifah, S.Pd. [Mahasiwa S2 Pendidikan Fisika]
Kang Marjan Fuadi Permadi, S.Pd. [Mahasiwa S2 Magister Pengajaran Fisika]
Kang Rimela Diaz, S.Pd. [Mahasiwa S2 Magister Pengajaran Fisika]

Anak-anakku di SMA Budi Luhur yang mengikuti SBMPTN dan Adik-adik di FORSALIM SMAN 1 Banjar Semoga Berhasil, raihlah cita-cita kalian.

Semangat!

Majulah Pendidikan Indonesia

Kita Bisa

Sumber: 

http://www.sbmptn.or.id/utama.php
Pikiran Rakyat
Republika Online
TELKOM Indonesia
Bank Mandiri

Semoga Bermanfaat

Sciences untuk Anak-Anak

Bagaimana Mengenalkan Sains Sejak Dini?

"Mari kita panggil Para Pemikir, Para Pemimpi, Para Perintis, Para Pencipta, Para Penemu, Para Penjelajah dan berikan mereka harapan, Life Starts Here"

*Nutrilon*

Anak memiliki rasa ingin tahu yang sangat tinggi. Rasa ingin tahu tersebut perlu difasilitasi oleh orang dewasa termasuk orang tua dan tenaga pendidik di dalamnya yang berfungsi sebagai guru anak. Anak dapat belajar apa saja asal tidak dipaksakan termasuk belajar sains sejak dini.

Belajar sains sejak dini dimulai dengan memperkenalkan alam dengan melibatkan lingkungan untuk memperkaya pengalaman anak. Anak akan belajar bereksperimen, bereksplorasi dan menginvestigasi lingkungan sekitarnya sehingga anak mampu membangun suatu pengetahuan yang nantinya dapat digunakan pada masa dewasanya.

Teori konstruktivis percaya bahwa pengetahuan akan dibangun secara aktif oleh anak melalui persepsi dan pengalaman langsung dengan lingkungannya. Anak yang banyak bersentuhan dengan alam akan lebih baik memaknai dunia mereka sehingga anak perlu mendapatkan kesempatan berinteraksi dengan lingkungan mereka yang akan membuat mereka secara aktif terus menerus mendapatkan pengetahuan.

Sciences for Kids

Math: Virtual Manipulatives NEW!!

Science Fairs

Mission to Mars TomatoSphere (A Mars-Related Project)

Space Shuttle Mission Profile

NEW! Healthy Kids Bullying NEW!!

Clipart Explorers Maps and Map Skills

Number the Stars Novel Study Homechild Novel Study

Night of the Twisters Novel Study

Hurricanes Tornadoes Clouds

Newspapers in Education

Math Story Problems Cartooning Writing Prompts Literature Circles
 
Reading Resources For Parents and Teachers

Anak-Anak Bermain di Taman yang Bersalju:

Belajar bersentuhan langsung dengan Alam.

Other Themes by Subject 

Earth Science Ecosystems Life Science Meteorology Physical Science Space Science Dragons

Archaeology Geography History Language Health Newfoundland

Ezines (Online Magazines) Arts Celebrations Reference Clipart

Dalam mendampingi dan memfasilitasi anak usia dini belajar sains, pendidik perlu memikirkan beberapa hal:

Pertama, apakah kita mengembangkan dan  menunjukkan sikap menghargai makhluk hidup?

Kedua, apakah pengalaman sains kita menekankan pada ketrampilan proses?

Misalnya mengamati, mengelompokkan, membandingkan, mengurutkan, meramalkan, mengkomunikasikan, mencoba, menduga.

Ketiga, apakah kegiatan sains kita masuk dalam kurikulum dan terintegrasi dengan area-area pembelajaran yang lain?

Keempat, apakah kita memberikan kesempatan dan bahan-bahan yang mendorong tiap anak untuk memanipulasi, mengeksplorasi, dan mengamati dengan menggunakan seluruh panca indra anak?

Misalnya: Apa yang kamu lihat?,

Apa yang kamu dengar?,

Seperti apa bentuknya?,

Apa yang dapat kamu bau?,

Seperti apa rasanya?,

apa saja yang kamu ingin ketahui dari benda itu?

Kelima, apakah kita mendorong pemikiran induktif dan deduktif ketika anak sedang bereksplorasi? Misalnya:

Apakah anak menggunakan fakta dan konsep untuk sampai pada kesimpulan umum?
Apakah anak dapat menduga fakta-fakta dan konsep-konsep khusus yang mendukung prinsip umum?

Semoga Bermanfaat

Maju Terus Pendidikan Indonesia

Sumber: 

Eli Tohonan Tua Pane, S.Pd.

NASA for Kids

Universitas Pendidikan Indonesia

Taman Pendidikan Al-Wustho

Apa Itu Angka Knudsen?

Didefinisikan sebagai rasio dari rata-rata panjang jalur bebas molekular terhadap suatu skala panjang fisik representatif tertentu. Skala panjang ini dapat berupa radius suatu benda dalam suatu fluida. 

Secara sederhana, angka Knudsen adalah berapa kali panjang diameter suatu partikel akan bergerak sebelum menabrak partikel lain. 

Intro:

The Knudsen number (Kn) is a dimensionless number defined as the ratio of the molecular mean free path length to a representative physical length scale. This length scale could be, for example, the radius of the body in a fluid. 

The number is named after Danish physicist Martin Knudsen (1871–1949). Who taught and conducted research at the Technical University of Denmark. He is primarily known for his study of molecular gas flow and the development of the Knudsen cell, which is a primary component of molecular beam epitaxy systems.

The Knudsen number is a dimensionless number defined as:

where

• = mean free path [L¹]
• = representative physical length scale [L¹].

For an ideal gas, the mean free path may be readily calculated so that:

where

• is the Boltzmann constant (1.3806504(24) × 10⁻²³ J/K in SI units), [M¹ L² T^-2 θ^-1]
• is the thermodynamic temperature, [θ¹]
• is the particle hard shell diameter, [L¹]
• is the total pressure, [M¹ L^-1 T^-2].

For particle dynamics in the atmosphere, and assuming standard temperature and pressure, i.e. 25 °C and 1 atm, we have ≈ 8 × 10⁻⁸ m.



Relationship to Mach and Reynolds numbers in gases


The Knudsen number can be related to the Mach number and the Reynolds number:
Noting the following:
Dynamic viscosity,

Average molecule speed (from Maxwell-Boltzmann distribution),

thus the mean free path,

dividing through by L (some characteristic length) the Knudsen number is obtained:

where

• is the average molecular speed from the Maxwell–Boltzmann distribution, [L¹ T^-1]
• T is the thermodynamic temperature, [θ¹]
• μ is the dynamic viscosity, [M¹ L^-1 T^-1]
• m is the molecular mass, [M¹]
• k_B is the Boltzmann constant, [M¹ L² T^-2 θ^-1]
• ρ is the density, [M¹ L^-3].

The dimensionless Mach number can be written:

where the speed of sound is given by

where

• U_∞ is the freestream speed, [L¹ T^-1]
• R is the Universal gas constant, (in SI, 8.314 47215 J K⁻¹ mol⁻¹), [M¹ L² T^-2 θ^-1 'mol'^-1]
• M is the molar mass, [M¹ 'mol'^-1]
• is the ratio of specific heats, and is dimensionless.

The dimensionless Reynolds number can be written:

Dividing the Mach number by the Reynolds number,

and by multiplying by ,

yields the Knudsen number.
The Mach, Reynolds and Knudsen numbers are therefore related by:

Aplikasi

The Knudsen number is useful for determining whether statistical mechanics or the continuum mechanics formulation of fluid dynamics should be used: If the Knudsen number is near or greater than one, the mean free path of a molecule is comparable to a length scale of the problem, and the continuum assumption of fluid mechanics is no longer a good approximation. In this case statistical methods must be used.

Problems with high Knudsen numbers include the calculation of the motion of a dust particle through the lower atmosphere, or the motion of a satellite through the exosphere. One of the most widely used applications for the Knudsen number is in microfluidics and MEMS device design. 

The solution of the flow around an aircraft has a low Knudsen number, making it firmly in the realm of continuum mechanics. Using the Knudsen number an adjustment for Stokes' Law can be used in the Cunningham correction factor, this is a drag force correction due to slip in small particles (i.e. d_p < 5 µm).

Semoga Bermanfaat. 

Ucapan Terima Kasih:

Bapak dan Ibu Guru Semasa SMA

Guru dan Dosen di Pendidikan Fisika, FPMIPA Universitas Pendidikan Indonesia

Sumber:

Arip Nurahman Notes

http://en.wikipedia.org/wiki/Knudsen_number

Fisika Super Hero

Sepanjang tahun dan bulan-bulan ini masyarakat dunia dihibur dengan tayangan film-film berbobot dari Hollywood, ada beberapa analisis dari penulis yang ingin mencoba menggambarkan sedikit opini dan riset kecil-kecilan.

Ada 4 buah Film yang sangat menarik perhatian penulis diantaranya adalah:

1. The Avengers
2. Men in Black 3
3. The Amazing Spider-Man 
4. The Dark Knight Rises

Tentu saja dalam berbagai film tersebut diceritakan kisah hidup manusia yang berputar pada permasalahan Cinta Kasih, Romansa, Persahabatan, Keluarga, Kemasyarakatan, Misteri, IPTEK bahkan Legenda.  

Sehingga mungkin saja kita bisa mengambil hikmah dan kebijaksanaan dari tayangan-tayangan tersebut namun kita juga harus berhati-hati karena sepertinya dalam film-film tersebut disisipkan beberapa ideologi dari kelompok-kelompok rahasia seperti dinyatakan dalam teori konspirasi. 

Dalam tulisan kali ini saya ingin berfokus pada perkembangan IPTEK wabil khusus Ilmu FISIKA yang ditunjukan dalam film-film tersebut.  Untuk masalah "perasaan" dalam film-film tersebut saya rasa  lebih baik pembaca nonton sendiri he.,he.,

1. The Avengers

 

Salah satu percakapan The Avengers yang melibatkan ilmu fisika.

Maria Hill   : "When did you become an expert in thermonuclear astrophysics?"

Tony Stark : "Last night."

Disamping itu juga ada terlihat kompleks penelitian NASA dalam film ini yang membahas mengenai DARK ENERGY dan DARK MATTER.

2. Men in Black 3

"The bitterest truth is better than the sweetest lie."

~Giffin~

Dalam film ini juga ada pembahasan mengenai perjalanan antar waktu yang melibatkan seorang "Makhluk Asing" bernama Giffin yang mampu melihat dimensi ke-5.

3. The Amazing Spider-Man 

 

Dr. Curt Connors: "If you want the truth, Peter, come and get it!"

Dilain cerita Peter Parker adalah seorang mahasiswa Fisika di Columbia University

Dalam film ini disajikan bagaimana sebuah penelitian dapat membuat penemuan-penemuan yang luar biasa seperti Biotechnology dan Genetics Engineering

4. The Dark Knight Rises

"A hero can be anyone. Even a man doing something as simple and reassuring as putting a coat around a little boy's shoulders to let him know that the world hadn't ended."

~Batman~

Dalam Film ini juga ada sebuah Reaktor Tenaga Fusi Nuklir yang menjadi Proyek Energi Rahasia Bruce Wayne (Batman) yang diilhami dari paper Ilmuwan Russia.

Akhirnya kita semua dapat melihat kehebatan sebuah karya orang-orang dan semoga kita yang menontonnya dapat mengambil pelajaran serta mengambil manfaatnya.

Ambil yang baiknya tinggalkan yang buruknya semoga!

Sumber:

1. The Avengers
2. Men in Black 3
3. The Amazing Spider-Man
4. The Dark Knight Rises

Fisika SMA

Pengalaman Belajar Fisika di SMAN BI 1 Banjar

Ada Apa Dengan Fisika?

 

 PAPER AIRPLANE ACTIVITY

Overview

In the paper airplane activity students select and build one of five different paper airplane designs and test them for distance and for time aloft. Part of this activity is designed to explore NASA developed software, FoilSim, with respect to the lift of an airfoil and the surface area of a wing.

Materials

• calculator
• ruler
• graph paper
• lightweight unlined paper
• scissors
• Download Design for the Dart
• Download FoilSim or execute on-line.

Technology Needed

• Internet Access
• Graphing Calculator (optional)

Time Required

• 2-3 class periods

Classroom Organization

• Students should work in groups of 3 or 4.

Procedure

1. Give students a sheet of unlined paper and instructions for construction of a paper airplane (See download above).
   
   
2. Students should give their plane a name using the aviation alphabet. (Example N 831 FE represents November 831 Foxtrot Echo. Identification numbers and letters must not exceed 7; and the identification must begin with N, which stands for the United States.)
   
   
3. Students should determine the area of the wings of their planes. If students are able, have them unfold their planes and lay out basic geometric shapes to fill the wing area. Then have them calculate the total area from the sum of the areas of the shapes. (See example. Use "back arrow" to return here.)
   
   If students are not able to calculate geometric areas, they could make a duplicate plane, cut off the wings, and lay the wings onto measured grids or pieces of graph paper and count the total squares covered, estimating partial squares.
   
   A variation of this technique that eliminates a duplicate plane and cutting wings is to draw or trace a grid on a blank transparency with a sharpie marker and then hold the clear grid over the wings to count squares covered.
   
   
4. Have students fly their planes in the gym or hallway or other large indoors area (to eliminate wind effects) five times, each time trying for maximum distance. Stress trying to duplicate the same launch angle and speed. Now do another five trials, this time trying for maximum time aloft. Students should record their distances and times and average the three longest distances and the three longest times.
   
   
5. Have students put their data onto a graph for the class, one graph of time aloft vs. wing area and the other of distance vs. wing area.
   
   
6. Discuss the results from the graphs as a class, and then ask for predictions as to what would happen if the wings were made smaller.
   
   
7. Have the students draw a line two centimeters from and parallel to the trailing edges of their wings, and then cut that 2 cm portion off the wings (Shown in red).
   
   The cut off part should be tucked on the inside of the plane when it is refolded in order to keep mass constant. You might ask the class to provide an explanation for doing this.
   
   
8. Repeat steps three through six.
   
   
9. Have the students investigate their results using FoilSim. They should set the ANGLE OF ATTACK to 5 degrees and then vary only the area of the wing and note the effect on the value of LIFT. They can compare these results to their own experimental results.
   
   
10. ADDITIONAL QUESTION: "Why don't all planes have the biggest wing area possible? Why do some fighter jets have small wings?" (ANSWER: There are other factors that contribute to lift, such as velocity and shape of the wing. The weight of a plane is also very important.) Students can investigate these other factors by going through the lessons that are part of FoilSim.

Extension Activity

• Further activities that your class might try.

Assessment Strategies/Evaluation

1. Each group could make a presentation on their airplane and what made its design successful.
2. Students could individually graph the experimental data and make a report.
3. Challenge students to fold a better plane and explain the reasons for changes in design.
4. Students could write a summary of experimental results and relate the variables tested.

Supplementary Resources

• Related Internet Links

Sumber:

NASA

Fisika SMA

Pengalaman Belajar Fisika di SMAN BI 1 Banjar

Ada Apa Dengan Fisika?

 Lift Formula

SUBJECT: Aeronautics
TOPIC: Lift
DESCRIPTION: A set of problems dealing with the aerodynamic lift equation.
CONTRIBUTED BY: Carol Hodanbosi
EDITED BY: Jonathan G. Fairman - August 1996

Purpose:

To understand for lift formula that determines aircraft lift capabilities.

Concept:

An aircraft's lift capabilities can be measured from the following formula:
L = (1/2) d v² s CL

• L = Lift, which must equal the airplane's weight in pounds
• d = density of the air. This will change due to altitude. These values can be found in a I.C.A.O. Standard Atmosphere Table.
• v = velocity of an aircraft expressed in feet per second
• s = the wing area of an aircraft in square feet
• CL = Coefficient of lift , which is determined by the type of airfoil and angle of attack.

The angle of attack and CL are related and can be found using a Velocity Relationship Curve Graph (see Chart B below).

Exercises

1. You are flying an F-117A fully equipped, which means that your aircraft weighs 52,500 pounds. You want to maintain equilibrium in straight and level flight at an altitude of 30,000 feet, cruising at 400 knots to conserve fuel. The aircraft's wing area is 1,140 square feet. At what angle of attack should the F-117A be set to maintain equilibrium? ** Additional information you need to know to solve this problem:
   How are knots related to mph to ft/s? See conversion below
   What is air density at 30,000 feet? See following chart (I.C.A.O.). chart A
   What is the relationship between CL and angle of attack?
   Use following Chart B to determine this value.
   
   1 knot = 1 nautical mile per hour = 6,076 ft per hour
   1 mph = 1 mile per hour = 5,280 feet per hour (answer)
   
2. To maintain an altitude of 45,000 feet at a constant speed requires an angle of attack of 4° . Your F-22 weighs 56,450 pounds, with a wing area of 830 square feet. To stay in level flight , what should your air speed indicate? The new F-22 air speed is indicated in Mach speed, not in knots. ** Additional information you need to know to solve this problem:
   How does Mach speed convert from feet per second or from knots?
   An angle of attack of 4° converts to what CL?
   What is the density of air at 45,000 feet?
   Mach 1 is the speed of sound, which varies with the air density. See the following I.C.A.O. chart (chart A) below. Mach 2 is twice the speed of sound, Mach 3 is three times the speed of sound.
   Use the Velocity Relationship Curve Graph chart below, Chart B, to find the appropriate CL from the given angle of attack.
   (answer)
   
3. At what velocity should you be traveling in the F-22 to maintain an 0° angle of attack at an altitude of 42,000 feet? Your weight is 57,600 pounds
   (answer)
4. Using the same plane (F-22) the same weight (57,600 pounds) and angle of attack, what should be its cruising speed at 36,000 feet? While descending from 42,000 feet to 36,000 feet the plane covers a horizontal distance of 40 nautical miles. What is the decent angle in degrees?
   (answer)

---

Chart A
I.C.A.O. Standard Atmosphere Table

   Altitude          Density     Speed of Sound
(Feet)             (d)         (Knots)
     0          .002377          661.7
 1,000          .002308          659.5
 2,000          .002241          657.2
 3,000          .002175          654.9
 4,000          .002111          652.6
 5,000          .002048          650.3
 6,000          .001987          647.9
 7,000          .001927          645.6
 8,000          .001868          643.3
 9,000          .001811          640.9
10,000          .001755          638.6
15,000          .001496          626.7
20,000          .001266          614.6
25,000          .001065          602.2
30,000          .000889          589.5
35,000          .000737          576.6
36,089*         .000706          573.8
40,000          .000585          573.8
45,000          .000460          573.8
50,000          .000362          573.8
55,000          .000285          573.8

* Geopotential of Tropopause

---

Chart B
Velocity Relationship Curve Graph

Fisika SMA

Pengalaman Belajar Fisika di SMAN BI 1 Banjar

Ada Apa Dengan Fisika?

Experiment on Fluids:
Finding the Velocity of a Fluid in a Confined Container

SUBJECT: Aeronautics
TOPIC: Fluid Velocity
DESCRIPTION: A set of mathematics problems dealing with fluid velocity.
CONTRIBUTED BY: Carol Hodanbosi
EDITED BY: Jonathan G. Fairman - August 1996

Purpose:

To calculate the velocity of a confined fluid, given the cross-section area and velocity of another region.

Concept:

The drawing below is a cross-section of a circular cone attached to a circular cylinder.

When a fluid (liquid or gas) is in a confined space, with no change in pressure or temperature, one can use the equation of continuity to find the velocity of the fluid if one knows the cross-section area and velocity in one of the regions. The formula for this is A1*V1 = A2*V2, where A is the cross-section area of one location and V is the velocity for that location.

Analysis:

Given three different locations in a confined container, A, B, and C, all having different radii, can you find the other two velocities of the fluid, if the velocity at A is given?

1. If the cross-section at A has a radius of 6 meters, can you find the area of the slice through the cone
   (Area = pi * r ²)?
   (answer)
2. If the velocity of the fluid at location A is 10.0 m/s, and the radius at location B is 4.2 meters, can you find the velocity at location B?
   (answer)
3. If the velocity at location C is 8.6 m/s, can you find the radius at location C?
   (answer) Extension:
4. If the radius of a fourth location, D, is one-half the radius of A , how would the velocity at location D compare to the velocity of the fluid at A? If D had one-third the radius of A, compare the velocity of the fluid at D to A. Explain your reasoning showing calculations.
   (answer)

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Fluids Pressure and Depth

SUBJECT: Aeronautics
TOPIC: Hydrostatic Pressure
DESCRIPTION: A set of mathematics problems dealing with hydrostatics.
CONTRIBUTED BY: Carol Hodanbosi
EDITED BY: Jonathan G. Fairman - August 1996

A fluid is a substance that flows easily. Gases and liquids are fluids, although sometimes the dividing line between liquids and solids is not always clear. Because of their ability to flow, fluids can exert buoyant forces, multiply forces in a hydraulic systems, allow aircraft to fly and ships to float.

The topic that this page will explore will be pressure and depth. If a fluid is within a container then the depth of an object placed in that fluid can be measured. The deeper the object is placed in the fluid, the more pressure it experiences. This is because is the weight of the fluid above it. The more dense the fluid above it, the more pressure is exerted on the object that is submerged, due to the weight of the fluid.

The formula that gives the P pressure on an object submerged in a fluid is:

P = r * g * h

where

• r (rho) is the density of the fluid,
• g is the acceleration of gravity
• h is the height of the fluid above the object

If the container is open to the atmosphere above, the added pressure must be included if one is to find the total pressure on an object. The total pressure is the same as absolute pressure on pressure gauges readings, while the gauge pressure is the same as the fluid pressure alone, not including atmospheric pressure.

P_total = P_atmosphere + P_fluid

P_total = P_atmosphere + ( r * g * h )

A Pascal is the unit of pressure in the metric system. It represents 1 newton/m²

Example:
Find the pressure on a scuba diver when she is 12 meters below the surface of the ocean. Assume standard atmospheric conditions.

Solution:
The density of sea water is 1.03 X 10 ³ kg/m³ and the atmospheric pressure is 1.01 x 10⁵ N/m².

P_fluid = r g h = (1.03 x10 ³ kg/m³) (9.8 m/s²) (12 m) = 1.21 x 10⁵ Newtons/m² P_total = P_atmosphere + P_fluid = (1.01 x 10⁵) + (1.21 x 10⁵ ) Pa = 2.22 x 10 ² kPa (kilo Pascals)

Exercises :

What is the pressure experienced at a point on the bottom of a swimming pool 9 meters in depth? The density of water is 1.00 x 10³ kg/m³.
(answer)

The interior of a submarine located at a depth of 45 meters is maintained at normal atmospheric conditions. Find the total force exerted on a 20 cm by 20 cm square window. Use the density of sea water given above.
(answer)

How many atmospheres is a depth of 100 meters of ocean water?
(answer)

If the weight density of pure water is 62 pounds/ft³, find the weight of water in a swimming pool whose dimensions are 20 ft by 10 ft by 6 feet.
(answer)

An airplane in level flight whose mass is 20,000 kg has a wing area of 60 m². What is the pressure difference between the upper and lower surfaces of its wing? Express your answer in atmospheres.
(answer)

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Pascal's Principle and Hydraulics

SUBJECT: Physics
TOPIC: Hydraulics
DESCRIPTION: A set of mathematics problems dealing with hydraulics.
CONTRIBUTED BY: Carol Hodanbosi
EDITED BY: Jonathan G. Fairman - August 1996

Hydraulic systems use a incompressible fluid, such as oil or water, to transmit forces from one location to another within the fluid. Most aircraft use hydraulics in the braking systems and landing gear. Pneumatic systems use compressible fluid, such as air, in their operation. Some aircraft utilize pneumatic systems for their brakes, landing gear and movement of flaps.

Pascal's law states that when there is an increase in pressure at any point in a confined fluid, there is an equal increase at every other point in the container.

A container, as shown below, contains a fluid. There is an increase in pressure as the length of the column of liquid increases, due to the increased mass of the fluid above.

For example, in the figure below, P3 would be the highest value of the three pressure readings, because it has the highest level of fluid above it.

If the above container had an increase in overall pressure, that same added pressure would affect each of the gauges (and the liquid throughout) the same. For example P1, P2, P3 were originally 1, 3, 5 units of pressure, and 5 units of pressure were added to the system, the new readings would be 6, 8, and 10.

Applied to a more complex system below, such as a hydraulic car lift, Pascal's law allows forces to be multiplied. The cylinder on the left shows a cross-section area of 1 square inch, while the cylinder on the right shows a cross-section area of 10 square inches. The cylinder on the left has a weight (force) on 1 pound acting downward on the piston, which lowers the fluid 10 inches. As a result of this force, the piston on the right lifts a 10 pound weight a distance of 1 inch.

The 1 pound load on the 1 square inch area causes an increase in pressure on the fluid in the system. This pressure is distributed equally throughout and acts on every square inch of the 10 square inch area of the large piston. As a result, the larger piston lifts up a 10 pound weight. The larger the cross-section area of the second piston, the larger the mechanical advantage, and the more weight it lifts.

The formulas that relate to this are shown below:

P1 = P2 (since the pressures are equal throughout).

Since pressure equals force per unit area, then it follows that

F1/A1 = F2/A2

It can be shown by substitution that the values shown above are correct,

1 pound / 1 square inches = 10 pounds / 10 square inches

Because the volume of fluid pushed down on the left side equals the volume of fluid that is lifted up on the right side, the following formula is also true.

V1 = V2

by substitution,

A1 D1 = A2 D2

• A = cross sectional area
• D = the distance moved

or

A1/A2= D2/D1

This system can be thought of as a simple machine (lever), since force is multiplied.The mechanical advantage can be found by rearranging terms in the above equation to

Mechanical Advantage(IMA) = D1/D2 = A2/A1

For the sample problem above, the IMA would be 10:1 (10 inches/ 1 inch or 10 square inches / 1 square inch).

Given these simple formulas, try to answer the questions below.

Exercises:

1. A hydraulic press has an input cylinder 1 inch in diameter and an output cylinder 6 inches in diameter.
   1. Assuming 100% efficiency, find the force exerted by the output piston when a force of 10 pounds is applied to the input piston.
      (answer)
      
      
   2. If the input piston is moved through 4 inches, how far is the output piston moved?
      (answer)
      
      
2. A hydraulic system is said to have a mechanical advantage of 40. Mechanical advantage (MA) is FR (output) / FE (input). If the input piston, with a 12 inch radius, has a force of 65 pounds pushing downward a distance of 20 inches, find
   
   
   1. the volume of fluid that has been displaced
      (answer)
      
      
   2. the upward force on the output piston
      (answer)
      
      
   3. the radius of the output piston
      (answer)
      
      
   4. the distance the output piston moves
      (answer)
      
      
      
3. What pressure does a 130 pound woman exert on the floor when she balances on one of her heels? Her heels have an average radius of 0.5 inch.
   (answer)
   
   
4. A car has a weight of 2500 pounds and rests on four tires, each having a surface area of contact with the ground of 14 square inches. What is the pressure the ground experiences beneath the tires that is due to the car?
   (answer)
   
   Extension :
   
5. The input and output pistons of a hydraulic jack are respectively 1 cm and 4 cm in diameter. A lever with a mechanical advantage of 6 is used to apply force to the input piston. How much mass can the jack lift if a force of 180 N is applied to the lever and efficiency is 80%?
   (answer)

Sumber:

NASA

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Buoyancy: Archimedes Principle

SUBJECT: Physics
TOPIC: Buoyancy
DESCRIPTION: A set of mathematics problems dealing with buoyancy.
CONTRIBUTED BY: Carol Hodanbosi
EDITED BY: Jonathan G. Fairman - August 1996

There are two types of flying machines that allow for lift to overcome gravity. The first type, called the aerodynamic machines such as helicopters and airplanes, rely on thrust and forward speed to produce lift. The second type, aerostatic machines, such as hot air balloons and lighter than air-type craft, rely on the differences in air density for lift.

This lesson is concerned with the second type, the type that are dependent on buoyancy.

If a cubic centimeter of aluminum was suspended in a fluid such as water with a very thin and negligible thread, the metal cube would have the fluid exerting pressure on the cube. Try to imagine that if the cube were to disappear, and the fluid would magically replace the cube, then the surrounding water would support this cube that is now containing water, so that the cube of water would be motionless. That is, the forces would be balanced. The cube of water would push out on the surrounding water and the surrounding water would push back on the cube. The fluid would be static, or stationary. Now replace this same cube of water with the original cube of aluminum. The surrounding water would not 'know' that the cube has been replaced with another substance. It would still push inward and upward and downward with the same force that it pushed on the cube of water. The sideways forces would be balanced and oppose each other equally, but the upward and downward forces would not be the same. The pressure at the bottom of the cube is greater than the pressure at the top of the cube, because pressure increases with increased depth. The difference between the upward and downward forces acting on the bottom and the top of the cube, respectively,is called buoyancy.

Using the aluminum as our example, it has a specific gravity of 2.8. Water has a specific gravity of 1.0. This means that a cubic centimeter of water would have a mass of 1.0 grams, while aluminum of the same size would have a mass of 2.8 grams. Since the aluminum cube displaces 1 cubic centimeter of water, it has a buoyancy of 1.0 grams. Since buoyancy is a force and not a mass, it must be converted to the proper units, which when multiplied by the acceleration of gravity (980 cm/s²) gives the units of dynes. That is,

(1.0 grams) (980 cm/s²) = 980 grams cm /s² or dynes

So our aluminum cube immersed in water would not 'weigh' (2.8 x 980) dynes or 2744 dynes. It would weigh less due to the fact it has a buoyant force of (1 x 980) dynes from the water. So it would weigh (2744-980) dynes or 1764 dynes while immersed in the water.

Archimedes Principle states that the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object.

Hot air balloons rise into the air because the density of the air (warmer air) inside the balloon is less dense than the air outside the balloon (cooler air). The balloon and the basket displaces a fluid that is heavier than the balloon and the basket, so it has a buoyant force acting on the system. Balloons tend to fly better in the morning, when the surrounding air is cool.

Exercises:

1. Find the weight of the air in a room with dimensions of 20 ft x 12 ft x 15 ft. The weight density of air at sea level is 0.08 pounds /ft³.
   (answer)
2. An iron anchor weighs 250 pounds in air and has a weight density of 480 lbs/ft³. If it is immersed in sea water that has a weight density of 64 lbs/ft³, how much force would be required to lift it while it is immersed?
   (answer)
3. An aluminum bar weighs 17 pounds in air. How much force is required to lift the bar while it is immersed in gasoline? The weight density of aluminum is 170 pounds /ft³ and that of gasoline is 42 pounds /ft³.
   (answer)
4. How much does a 20 ft x 10 ft x 8 ft swimming pool filled with water weigh? Assume the water has a weight density of 62 lbs/ft³.
   (answer)
5. A balloon weighing 80 kg has a capacity of 1200 m³. If it is filled with helium, how great a payload can it support? The density of helium is 0.18 kg/ m³ and the density of air is 1.30 kg/ m³. Express your answer in newtons.
   (answer)

 Sumber:

NASA

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Newton's Third Law of Motion

SUBJECT: Physics
TOPIC: Force and Motion
DESCRIPTION: A set of mathematics problems dealing with Newton's Laws of Motion.
CONTRIBUTED BY: Carol Hodanbosi
EDITED BY: Jonathan G. Fairman - August 1996

The third law of motion states that if a body exerts a force on a second body, the second body exerts a force that is equal in magnitude and opposite in direction to the first force. So for every action force there is always a reaction force. No force can occur by itself

The book lying on the table is exerting a downward force on the table, while the table is exerting an upward reaction force on the book. Because the forces are equal and opposite, the book remains at rest. Notice also that the table legs are in contact with the floor and exert a force downward on it, while the floor in turn exerts an equal and opposite force upward.

Questions for you to consider:

If forces are always equal and opposite in action and reaction, how is it possible for an object to accelerate?
(Answer)

Explain, in detail, using the third law of motion, how a person is able to walk forward.
(Answer)

There is a classic problem that physicists like to ask students. A horse is pulling a carriage on a level ground. The horse knows the third law of motion. He tells the carriage that he will exert a force forward, and the carriage will exert a force equal to the horse's force but in opposite directions. Therefore, the horse explained, he can never pull the carriage forward. Can you explain to the horse that he is mistaken? How is he able to pull the carriage forward?
(Answer)

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The First and Second Laws of Motion

SUBJECT: Physics
TOPIC: Force and Motion
DESCRIPTION: A set of mathematics problems dealing with Newton's Laws of Motion.
CONTRIBUTED BY: Carol Hodanbosi
EDITED BY: Jonathan G. Fairman - August 1996

Newton's First Law of Motion states that a body at rest will remain at rest unless an outside force acts on it, and a body in motion at a constant velocity will remain in motion in a straight line unless acted upon by an outside force.

If a body experiences an acceleration ( or deceleration) or a change in direction of motion, it must have an outside force acting on it. Outside forces are sometimes called net forces or unbalanced forces.

The property that a body has that resists motion if at rest, or resists speeding or slowing up, if in motion, is called inertia. Inertia is proportional to a body's mass, or the amount of matter that a body has. The more mass a body has, the more inertia it has.

The Second Law of Motion states that if an unbalanced force acts on a body, that body will experience acceleration ( or deceleration), that is, a change of speed. One can say that a body at rest is considered to have zero speed, ( a constant speed). So any force that causes a body to move is an unbalanced force. Also, any force, such as friction, or gravity, that causes a body to slow down or speed up, is an unbalanced force. This law can be shown by the following formula

F= ma

• F is the unbalanced force
• m is the object's mass
• a is the acceleration that the force causes

If the units of force are in newtons, then the units of mass are kilograms and the units of acceleration are m/s². If the units of force are in pounds (English), then the units of mass are in slugs, and the units of acceleration are ft/s².

Motion of an object that is not accelerated (moving at a constant speed and in a straight line) can be found using the formula

d= v t

• d is the distance traveled
• v is the rate of motion (velocity)
• t is the time

Some sample problems that illustrates the first and second laws of motion are shown below:

Example 1
If the speed of sound on a particular day is 343 m/s, and an echo takes 2.5 seconds to return from a cliff far away, can you determine how far the cliff is from the person making the sound?

An echo is a sound that travels out and back. It take 2.5 seconds for this trip, which is twice the distance to the cliff. Therefore, it only takes 1.25 seconds for the sound to reach the cliff. By substitution,

d = v t
d = (343 m/s) (1.25 s)
d = 429 m

Example 2
If an unbalanced force of 600 newtons acts on a body to accelerate it at +15 m/s², what is the mass of the body?

F = ma
m=F/a
m = 600n/15 m/s2
m= 40 kg

Exercises:

1. If a car is traveling at 50 km/hr along a straight line, how many meters does it travel in 10 seconds?
   (Answer)
2. A force of 5000 newtons is applied to a 1200 kg car at rest. What is its acceleration?
   (Answer)
3. A 10 kg body has an acceleration of 2 m/s². Find the net force acting on the body.
   (Answer)
4. An empty truck with a mass of 2500 kg has an engine that will accelerate at a rate of 1.5 m/s². What will be the acceleration when the truck has an additional load of 1500 kg ?
   (Answer)
5. A box resting on a table has a mass of 5.0 kg.
   1. What is its weight?
   2. What will be its acceleration when an unbalanced horizontal force of 40 newtons acts on it?
      (Answer)
      
      
6. What is the weight of an object that has a mass of 60 slugs?
   (Answer)
7. A net force of 75 pounds acts on a body of 25 slugs. The body is initially at rest. What is its acceleration during the action of the force?
   (Answer)
8. What is the mass of a 185 pound man? If a 100 pound horizontal net force acts on the man while he is sitting on a wooden floor, what will his acceleration be?
   (Answer)