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Inessa [10]
2 years ago
11

A 5⁢kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant.

The mass of the planet is unknown. After 2⁢s, the object has fallen 30m. Air resistance is considered to be negligible. What is the gravitational force exerted on the 5kg5⁢kg object near the planet’s surface?
(A) 5 N
(B) 15 N
(C) 37.5 N
(D) 75 N
Physics
1 answer:
Umnica [9.8K]2 years ago
8 0

Answer:

The gravitational force exerted on the object is 75 N (answer D)

Explanation:

Hi there!

The gravitational force is calculated as follows:

F = m · g

Where:

F = force of gravity.

m = mass of the object.

g = acceleration due to gravity (unknown).

For a falling object moving in a straight line, its height at a given time can be calculated using the following equation:

y = y0 + v0 · t + 1/2 · a · t²

Where:

y = position at time t.

y0 = initial position.

v0 = initial velocity.

t = time.

g = acceleration due to gravity.

Let´s place the origin of the frame of reference at the point where the object is released so that y0 = 0. Let´s also consider the downward direction as negative.

Then, after 2 seconds, the height of the object will be -30 m:

y = y0 + v0 · t + 1/2 · g · t²

-30 m = 0 m + 0 m/s · 2 s + 1/2 · g · (2 s)²

-30 m = 1/2 · g · 4 s²

-30 m = 2 s ² · g

-30 m/2 s² = g

g = -15 m/s²

Then, the magnitude of the gravitational force will be:

F = m · g

F = 5 kg · 15 m/s²

F = 75 N

The gravitational force exerted on the object is 75 N (answer D)

Have a nice day!

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Question 1
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Answer:

1)  g = 4π² / m, 3) xaxis the  length of the pendulums and the y axis the period squared

Explanation:

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         w = √ g / l

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         w = 2π f = 2π / T

we substitute

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with this equation they can determine the value of the acceleration of gravity, for this they measure the period for various lengths of the pendulum and graph

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where this is the equation of a line if the independent variable is y = T² and x = l

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so the slope is

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clearing

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where the slope can be found with the values ​​of the line not the experimental values.

2) to carry out the experiment, or the thread is attached to the sphere, the length of the pendulum that goes from the pivot point to the center of the sphere is measured with a tape measure and a small finished angle is turned or less than 10th is released, it is good to wait for the first oscillation to walk, the time of a determined number of oscillations is generally measured 10 or 20, the period is calculated

    T = t / n

a table of T² against the length is made and it is plotted with the length in the ax ax, we look for the slope and hence the acceleration of gravity

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4) of the equation of the line

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                 where it ends up reaching the floor

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5) when the spring is cut, the sphere remains under the effect of gravity acceleration, the harmonic movement disappears and the sphere is in a vertical movement

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