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AnnyKZ [126]
3 years ago
10

the sputnik 1 satellite orbited earth (mass=5.98 x 10^24 kg) in a circle of radius 6.96 x 10^6 m. what was its orbital velocity?

Physics
2 answers:
user100 [1]3 years ago
8 0

Answer:

v = 4.79 10³ m / s

Explanation:

For this exercise we will use the universal law of gravitation and Newton's second law

          F = G M m / r²

where G is the gravitational constant, m the mass of the satellite, M the mass of the Earth, r the distance from the center of the planet to the satellite

           F = m a

since the satellite has a circular path, the acceleration is centripetal

          a = v² / r

we substitute

          G M m / r² = m v² / r

           G M / r = v²

   

We calculate

          v² = 6.67 10⁻¹¹ 5.98 10²⁴ / 6.96 10⁶

          v = √ (22.94 10⁶)

          v = 4.79 10³ m / s

galben [10]3 years ago
7 0

Answer:

7570 m/s

Explanation:

take the square root of: 6.67E-11× 5.98E24/6.96E6

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A charged box ( m = 495 g, q = + 2.50 μ C ) is placed on a frictionless incline plane. Another charged box ( Q = + 75.0 μ C ) is
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Answer:

a=16.2m/s^{2}

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When fixed,F=o

Hence

masin\alpha =\frac{kQq}{r^{2}}\\0.495kg*asin35=\frac{9*10^{9}*2.5*10^{-6}*75.0*10^{-6}}{0.61^{2}} \\0.28a=4.5351\\a=\frac{4.5351}{0.28}\\\\ a=16.2m/s^{2}

The value of the acceleration is 16.2m/s^2

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Forces that act on an object but are not equal in size
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Leo places a plant in front of the center of curvature of a concave mirror. Which characteristics will the image of the plant ha
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Read 2 more answers
3. If the distance of the screen is moved from 100. cm to 200. cm the area of light would in-crease from 150. cm^2 to
german

Answer:

3) C

4 D

5) C

Explanation:

3) given that

Initial distance of the screen = 100cm

Initial area = 150 cm^2

Final distance = 200 cm

The intensity of light is inversely proportional to the square of the distance. That is

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And also I = P/A

1/d^2 = P/A

P = A/d^2

P1 = P2

150/100 = A/200

1.5 = A/200

A = 1.5 × 200

A = 300 cm^2

4.) Light is projected onto a screen 75.0 cm from a light source. The light intensity = 4436 lux

If the screen is moved from 75.0 cm to 150. cm, the light sensor reading will be

Using inverse square law

I = 1/d^2

I×d^2 = constant. Therefore,

4436 × 75^2 = I × 150^2

I = 24952500/22500

I = 1109 lux

5.) We can express the relationship between luminosity, brightness, and distance with a simple formula.

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