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irakobra [83]
3 years ago
13

Geosynchronous satellites orbit the Earth at a distance of 42 000 km from the Earth's center. Their angular speed at this height

is the same as the rotation rate of the Earth, so they appear stationary at certain locations in the sky. What is the force acting on a 1 500-kg satellite at this height?
Physics
1 answer:
frozen [14]3 years ago
7 0

Answer:

Force will be 7.97 N

Explanation:

We have given that earth is at s distance of 42000 km from the earth center

So r = 42000 km = 42000000 m

Mass m = 1500 kg

Angular velocity of earth \omega =7.29\times 10^{-5}rad/sec

We have to find the force

We know that force is given by

F=m\omega ^2r=1500\times (7.29\times 10^{-5})^2\times 42000000=797.1615\times 10^{-2}=7.97N

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Vikentia [17]
Mutation because they change the genes
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3 years ago
A spring stretches 5 cm when a 300-N mass is suspended from it. Calculate the spring constant in N / m .
rusak2 [61]

Answer:

Spring constant in N / m = 6,000

Explanation:

Given:

Length of spring stretches = 5 cm = 0.05 m

Force = 300 N

Find:

Spring constant in N / m

Computation:

Spring constant in N / m = Force/Distance

Spring constant in N / m = 300 / 0.05

Spring constant in N / m = 6,000

8 0
3 years ago
A substance did not change its chemical nature in a reaction. Which most likely describes the reaction?
Andre45 [30]

B. It was hit with a hammer.

4 0
3 years ago
Read 2 more answers
A cat runs at 2 m/s for 3 s and then slows to a stop with an acceleration of 0.80 m/s2. What is
sp2606 [1]

Answer:

see that the correct one is B

Explanation:

To solve this exercise let us use the kinematic relations

           v² = v₀² - 2 a x

as they indicate that the car stops, therefore the final speed is yield v = 0

          x = v₀² / 2a

let's calculate

          x = 2²/(2 0.8)

         x = 2.5 m / s²

When reviewing the answers we see that the correct one is B

7 0
3 years ago
What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL
lozanna [386]

Answer:

\displaystyle \Delta x=1.74\ m

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

\displaystyle PE = \frac{1}{2}k(\Delta x)^2

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

\displaystyle \Delta x=\swrt{\frac{2PE}{k}}

Substituting:

\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}

Calculating:

\displaystyle \Delta x=\sqrt{3.0175}

\boxed{\displaystyle \Delta x=1.74\ m}

6 0
3 years ago
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