According to Newton’s law of universal gravitation, which statement is true?

Physics

2 answers:

olasank [31]2 years ago

6 0

A) Falling objects accelerate as they approach the ground.

True

Reason : as the falling object approach the ground, acceleration due to gravity in down direction acts on it and accelerate it.

B) The weight of an object increases as it moves away from the ground.

False

Reason : As the object moves away from the ground, acceleration due to gravity acting on the object decreases. since the weight directly depends on the acceleration due to gravity, hence the weight decreases too.

C) Lighter objects experience a lower acceleration than heavier objects.

False

Reason : Since acceleration due to gravity is independent of the mass.

D) The mass of an object increases as it moves closer to the ground.

False

Reason : Since mass of an object is independent of the acceleration due to gravity.

SIZIF [17.4K]2 years ago

4 0

Option (a) is correct.

Falling objects accelerate as they approach the ground.This is because of the force of gravity acting on the falling objects. so the velocity of these objects increases continuously as they approach the ground. the acceleration acting on the falling objects is a constant ( close to the surface of earth) and is called as acceleration due to gravity denoted by g. value of g=9.8 m/s².

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vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses

dsp73

Answer:

The effective spring constant of the firing mechanism is 1808N/m.

Explanation:

First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

$v_0_x=\frac{x}{t}\\ \\v_0\cos\theta=\frac{x}{t}\\\\v_0=\frac{x}{t\cos\theta}$

(This is correct because the horizontal motion has acceleration zero). Then:

$v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s$

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

$E_0=E_f\\\\U_e=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\\implies k=\frac{mv^2}{x^2}$