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3 years ago

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

Physics

2 answers:

olasank [31]3 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]3 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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A ball is thrown straight upward. How does the sign of the work done by gravity while the ball is traveling upward compare with

Vlada [557]

If I’m understanding this. Th reason why the ball can resist gravity’s pull is because the force used to throw it up was greater than gravity’s pull on it. BTW, gravity’s pull is 9.8m/s. So it would have to be greater than that

8 0

3 years ago

Read 2 more answers

an object is shot vertically upward into the air with an initial velocity of 20 m/s? which of the following correctly describes

Drupady [299]

Answer:

Answer is (C)

Explanation:

For acceleration

• If the motion is vertically, then acceleration is 9.8 m/s²

» Upward motion, acceleration is negative (-9.8)

» Downward motion, acceleration is positive (+9.8)

For velocity

${ \rm{v = u + gt}} \\ { \rm{v = 20 + ( - 9.8 \times 4)}} \\ { \rm{v = - 19.2 \: m {s}^{ - 1} }}$

5 0

2 years ago

Can someone please answer this for me

UNO [17]

Because in my opinion they do bending of a light wave as it passes at an angle from one medium to another.

5 0

3 years ago

At one instant a bicyclist is 21.0 m due east of a park's flagpole, going due south with a speed of 13.0 m/s. Then 21.0 s later,

andre [41]

Answer:

Part a)

$d = 21\sqrt2 = 29.7 m$

Part b)

Direction is 45 degree North of West

Part c)

$v_{avg} = 1.41 m/s$

Part d)

direction of velocity will be 45 Degree North of West

Part e)

$a = 0.875 m/s^2$

Part f)

$\theta = 45 degree$ North of East

Explanation:

Initial position of the cyclist is given as

$r_1 = 21.0 m$ due East

final position of the cyclist after t = 21.0 s

$r_2 = 21.0 m$ due North

Part a)

for displacement we can find the change in the position of the cyclist

so we have

$d = r_2 - r_1$

$d = 21\hat j - 21\hat i$

so magnitude of the displacement is given as

$d = 21\sqrt2 = 29.7 m$

Part b)

direction of the displacement is given as

$\theta = tan^{-1}\frac{y}{x}$

$\theta = tan^{-1}\frac{21}{-21}$

so it is 45 degree North of West

Part c)

For average velocity we know that it is defined as the ratio of displacement and time

so here the magnitude of average velocity is defined as

$v_{avg} = \frac{\Delta x}{t}$

$v_{avg} = \frac{29.7}{21}$

$v_{avg} = 1.41 m/s$

Part d)

As we know that average velocity direction is always same as that of average displacement direction

so here direction of displacement will be 45 Degree North of West

Part e)

Here we also know that initial velocity of the cyclist is 13 m/s due South while after t = 21 s its velocity is 13 m/s due East

So we have

change in velocity of the cyclist is given as

$\Delta v = v_f - v_i$

$\Delta v = 13\hat i - (-13\hat j)$

now average acceleration is given as

$a = \frac{\Delta v}{\Delta t}$

$a = \frac{13\hat i + 13\hat j}{21}$

so the magnitude of acceleration is given as

$a = \frac{13\sqrt2}{21} = 0.875 m/s^2$

Part f)

direction of acceleration is given as

$\theta = tan^{-1}\frac{y}{x}$

$\theta = tan^{-1}\frac{13}{13}$

$\theta = 45 degree$ North of East

4 0

3 years ago

Which of the following contains the most thermal energy?

Evgesh-ka [11]

c. a hot tub full of hot water

5 0

4 years ago