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timofeeve [1]
3 years ago
8

Radio signals travel at a rate of 3 × 108 meters per second. how many seconds would it take for a radio signal to travel from a

satellite to the surface of earth if the satellite is orbiting at a height of 9.6 × 106 meters? (hint: time is distance divided by speed.)
Physics
1 answer:
nlexa [21]3 years ago
6 0
<span>3.2x10^-2 seconds (0.032 seconds)
   This is a simple matter of division. I also suspect it's an exercise in scientific notation, so here is how you divide in scientific notation:

   9.6 x 10^6 m / 3x10^8 m/s

   First, divide the significands like you would normally.
 9.6 / 3 = 3.2

   And subtract the exponent. So
 6 - 8 = -2

   So the answer is 3.2 x 10^-2
 And since the significand is less than 10 and at least 1, we don't need to normalize it.

   So it takes 3.2x10^-2 seconds for the radio signal to reach the satellite.</span>
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The radius of a planet is 2400 km, and the acceleration due to gravity at its surface is 3.6 m/s2.
kiruha [24]

Answer:

3.1\cdot10^{23}\:\mathrm{kg}

Explanation:

We can use Newton's Universal Law of Gravitation to solve this problem:

g_P=G\frac{m}{r^2}., where g_P is acceleration due to gravity at the planet's surface, G is gravitational constant 6.67\cdot 10^{-11}, m is the mass of the planet, and r is the radius of the planet.

Since acceleration due to gravity is given as m/s^2, our radius should be meters. Therefore, convert 2400 kilometers to meters:

2400\:\mathrm{km}=2,400,000\:\mathrm{m}.

Now plugging in our values, we get:

3.6=6.67\cdot10^{-11}\frac{m}{(2,400,000)^2},

Solving for m:

m=\frac{2,400,000^2\cdot3.6}{6.67\cdot 10^{-11}},\\m=\fbox{$3.1\cdot10^{23}\:\mathrm{kg}$}.

6 0
2 years ago
What is escape velocity?
olga_2 [115]
<span>It is the lowest velocity which a body must have in order to escape the gravitational attraction of a particular planet or other object.

Every planet has their own corresponding escape velocities. Example - Earth has escape velocity of 11.2 Km/s. It means, if you want to leave the Earth's gravitational field then it's the lowest speed which you need to acquire otherwise you wouldn't do that!

Hope this helps!</span>
6 0
2 years ago
Suppose that a merry-go-round, which can be approximated as a disk, has no one on it, but it is rotating about a central vertica
seropon [69]

Answer:

0.1 rev/s

Explanation:

M = mass of the merry go round = 200 kg

R = radius of merry go round = 6 m

I_{o} = Moment of inertia of merry go round = (0.5) MR² = (0.5) (200) (6)² = 3600 kgm²

m = mass of the man = 100 kg

I_{f} = Moment of inertia of merry go round when man sits on it at the edge = (0.5) MR² + mR² = (0.5) (200) (6)² + (100) (6)² = 7200 kgm²

w_{o} = initial Angular speed of merry-go-round before man sit = 0.2 rev/s

w_{f} = Angular speed of merry-go-round after man sit = ?

Using conservation of angular momentum

I_{o} w_{o} = I_{f} w_{f}

(3600) (0.2) = (7200) w_{f}

w_{f} = 0.1 rev/s

4 0
3 years ago
What type of device is a refrigerator?
Tamiku [17]
A heat pump that uses work to move heat
7 0
2 years ago
The nucleus of the polonium isotope 214 Po (mass 214 u) is radioactive and decays by emitting an alpha particle (a helium nucleu
amid [387]

Answer:

368224.29906 m/s

Explanation:

M = Mass of Polonium nucleus = 214 u

V = Velocity of nucleus

m = Mass of Helium nucleus = 4 u

v = Velocity of alpha particle = 1.97\times 10^7\ m/s

In this system the momentum is conserved

MV+mv=0\\\Rightarrow MV=-mv\\\Rightarrow 214V=-4\times 1.97\times 10^7\\\Rightarrow V=\dfrac{-4\times 1.97\times 10^7}{214}\\\Rightarrow V=368224.29906\ m/s

The recoil speed of the nucleus that remains after the decay is 368224.29906 m/s

6 0
3 years ago
Read 2 more answers
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