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Verdich [7]
3 years ago
13

Planet Kling has half the radius and 2 times the mass of the Earth. What is the best estimate for the magnitude of the gravitati

onal field at the surface of planet Kling?
Physics
2 answers:
Vlad [161]3 years ago
6 0

Answer:80 m/s^2

Explanation:

alekssr [168]3 years ago
3 0

Answer:

80m/s²

Explanation:

First, use the equation below to find the strength of a planet/moon's satellite field and acceleration rate.

This is G × Mass of planet/moon ÷ r² (which is the distance or conversely radius squared)

Then you want to plg in all your values, which are going to be the gravitational constant, Kling's mass, and Kling's radius.

Remember that Earth's mass is equal to 5.97 × 10∧24 kg and its radius is exactly 6,731 km, so this will help you figure out your values for any other problems like this .

Also that the big ( G) you see is = 6.673×10∧-11.

Now for step two, which is the last thing you need to do. Once you've simplified the above equation, take the value you found, which should ≈ 7.85, and then multiply it by the force of gravity on Earth.

Once you multiply 7.85 by 9.8 m/s², the answer will be = 76.93 or ≈ 80m/s².

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A ball is dropped from rest at point O. After falling for some time, it passes by a window of height 3.3 m and it does so in 0.2
stiv31 [10]

Answer:

Speed at which the ball passes the window’s top = 10.89 m/s

Explanation:

Height of window = 3.3 m

Time took to cover window = 0.27 s

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We have equation of motion s = ut + 0.5at²

For the top of window (position A)

                     s_A=0\times t_A+0.5\times 9.81t_A^2\\\\s_A=4.905t_A^2

For the bottom of window (position B)

                     s_B=0\times t_B+0.5\times 9.81t_B^2\\\\s_A=4.905t_B^2

\texttt{Height of window=}s_B-s_A=3.3\\\\4.905t_B^2-4.905t_A^2=3.3\\\\t_B^2-t_A^2=0.673

We also have

                 t_B-t_A=0.27

Solving

         t_B=0.27+t_A\\\\(0.27+t_A)^2-t_A^2=0.673\\\\t_A^2+0.54t_A+0.0729-t_A^2=0.673\\\\t_A=1.11s\\\\t_B=0.27+1.11=1.38s

So after 1.11 seconds ball reaches at top of window,

       We have equation of motion v = u + at

                                     v_A=0+9.81\times 1.11=10.89m/s

Speed at which the ball passes the window’s top = 10.89 m/s                

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