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jekas [21]
3 years ago
15

An auto, starting from rest, undergoes constant acceleration and covers a distance of 1200 meters. The final speed of the auto i

s 60 meters/sec. How long does it take the car to cover the 1200 meters?
Physics
1 answer:
Elanso [62]3 years ago
8 0

Answer:

It will take 40 sec

Explanation:

We have given that auto is starting from rest so initial velocity u = 0 m/sec

And final velocity is given as v = 60 m/sec

Distance traveled by car s = 1200 m

According third equation of motion we know that v^2=u^2+2as

So 60^2=0^2+2\times a\times 1200

a=1.5m/sec^2

Now according to first equation of motion v = u+at

So 60=0+1.5\times t

t = 40 sec

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Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<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:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



M=1.9(10^{27})kg is the mass of Jupiter


a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, 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:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



Then:


<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



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A hopper jumps straight up to a height of 1.3 m. With what velocity did he leave the floor
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The velocity with which the jumper leaves the floor is 5.1 m/s.

<h3>What is the initial velocity of the jumper?</h3>

The initial velocity of the jumper or the velocity with which the jumper leaves the floor is calculated by applying the principle of conservation of energy as shown below.

Kinetic energy of the jumper at the floor = Potential energy of the jumper at the maximum height

¹/₂mv² = mgh

v² = 2gh

v = √2gh

where;

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  • g is acceleration due to gravity

v = √(2 x 9.8 x 1.3)

v = 5.1 m/s

Learn more about initial velocity here: brainly.com/question/19365526
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3 0
11 months ago
Which of the following is an example of acceleration?
Naddika [18.5K]

Answer:

c

Explanation:

When a satellite is orbiting the earth , a constant force is being applied on it which means it must has acceleration. Also the direction of satellite is always being changed when it is orbitting to there is always change in the velocity vector which means acceleration.

You can view in the attached diagram to understand how the velocity is being changed.

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