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3 years ago

What is the force applied to a baseball that has a mass of 142kg and has a acceleration of 30m/s to the power of 2

Physics

1 answer:

aev [14]3 years ago

7 0

Mass=m=142kg
Acceleration=a=30m/s
Force=F

Using Newton's second law

$\\ \sf\Rrightarrow F=ma$

$\\ \sf\Rrightarrow F=142(30)$

$\\ \sf\Rrightarrow F=4260N$

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If earth's mass were half its actual value but its radius stayed the same, the escape velocity of earth would be:________

siniylev [52]

If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be $V_e = \sqrt{\dfrac{GM}{r}}$ .

<h3>What is an escape velocity?</h3>

The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.

The formula to calculate the escape velocity of earth is given below:-

$V_e=\sqrt{\dfrac{2GM}{r}}$

Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-

$V_e=\sqrt{\dfrac{2GM}{r\times 2}}$

$V_e = \sqrt{\dfrac{GM}{r}}$ .

Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be $V_e = \sqrt{\dfrac{GM}{r}}$ .

To know more about escape velocity follow

brainly.com/question/14042253

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1 year ago

A person is attracted toward the centerof the earth by a 500 n gravitational force. the force with which the earth is attracted

nignag [31]

500 N is the answer, you just tell that the moon is attracted towards the person because of the Earth's huge mass.

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3 years ago

you stop the stopwatch at 4.0 s, but you notice a short time later that the same ant is at 0.81 m on the meter stick. Assuming t

telo118 [61]

The time elapsed since you stopped the stopwatch is 0.41 s.

<em>Your question is not complete, it seems to be missing the following information;</em>

"The velocity of the ant is 2 m/s"

The given parameters;

velocity of the ant, v = 2 m/s

change in position of the ant, Δx = 0.81 m

initial time, t₁ = 4 s

time when the ant was noticed, = t₂

Velocity is defined as the change in displacement per change in time of motion of an object.

$v = \frac{\Delta x}{\Delta t} = \frac{\Delta x}{t_2 - t_1} \\\\t_2 -t_1 = \frac{\Delta x}{v} \\\\t_2 - 4 = \frac{0.81}{2} \\\\t_2 - 4 = 0.405\\\\t_2 = 0.405 + 4\\\\t_2 = 4.405 \approx 4.41 \ s$

The time elapsed since you stopped the stopwatch is calculated as;

$t_{elapsed} = 4.41 \ s - 4\ s = 0.41 \ s$

Thus, the time elapsed since you stopped the stopwatch is 0.41 s.

Learn more here: brainly.com/question/18153640

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