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Bogdan [553]
3 years ago
6

Use the impulse-momentum theorem to find how long a stone falling straight down takes to increase its speed from 3.7 m/s to 9.90

m/s .
Physics
1 answer:
11111nata11111 [884]3 years ago
8 0
<h2>Time taken is 0.632 seconds</h2>

Explanation:

Impulse momentum theorem is change in momentum is impulse.

Change in momentum = Impulse

Final momentum - Initial momentum = Impulse

Mass x Final velocity - Mass x Initial Velocity = Force x Time

Mass x Final velocity - Mass x Initial Velocity =Mass x Acceleration x Time

Final velocity - Initial Velocity = Acceleration x Time

Final velocity = 9.9 m/s

Initial Velocity = 3.7 m/s

Acceleration = 9.81 m/s²

Substituting

9.9 - 3.7 = 9.81 x Time

Time = 0.632 seconds

Time taken is 0.632 seconds

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coldgirl [10]

Answer:

<h3>The answer is 8.5 kg</h3>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

where

f is the force

a is the acceleration

So we have

m =  \frac{34}{4}  =  \frac{17}{2}  \\

We have the final answer as

<h3>8.5 kg</h3>

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Answer:

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What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

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M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

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