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3 years ago

Which is an effect of gravity on objects on the surface of Earth? Check all that apply.

Physics

1 answer:

timama [110]3 years ago

3 0

• Objects “fall” toward the center of Earth. (TRUE)

Since the gravitational force between earth and any object is always along the line joining the centers of two so all the objects near the surface of earth will always feel force towards the centre of the earth due to gravity.

• Objects are pulled by Earth but do not pull on Earth. (FALSE)

Since gravitational force will always follow Newton's III law as per which every action has equal and opposite reaction force so object near the surface of earth will also pull the earth as earth is pulling the object towards the center.

• Objects accelerate at a rate of 9.8 m/s each second. (TRUE)

Since acceleration due to gravity of Earth is 9.8 m/s^2 so here we can say that all objects near the surface of earth will accelerate towards the center of earth at rate of 9.8 m/s^2 due to gravitational pull.

• Objects travel at a rate of 9.8 m/s to are the center of Earth. (FALSE)

Since earth is pulling all objects near its surface with gravitational force so objects can not move at constant speed but it will move with constant acceleration.

• Objects are pulled by Earth more strongly than they pull on Earth. (FALSE)

Since gravitational force will always follow Newton's III law which say every action has equal and opposite reaction so we can say that force due to all objects on earth will be same as the force of earth on the object.

You might be interested in

Why does the value of g vary from place to place?

NemiM [27]

Answer:

The variation in apparent gravitational acceleration (g) at different locations on Earth is caused by two things (as you implied). ... The distance between the centers of mass of two objects affects the gravitational force between them, so the force of gravity on an object is smaller at the equator compared to the poles.

8 0

3 years ago

6. The hole on a level, elevated golf green is a horizontal distance of 150 m from the tee and at an elevation of 12.4 m above t

Georgia [21]

Answer:

u = 104.68 m/s

Explanation:

given,

horizontal distance = 150 m

elevation of 12.4 m

angle = 8.6°

horizontal motion = x = u cos θ. t .............(1)

vertical motion =

$y = u sin \theta - \dfrac{1}{2}gt^2$ ................(2)

from equation(1) and (2)

$y = x tan \theta - \dfrac{gx^2}{2u^2cos^2\theta}$ ..........{3}

$12.4 = 150\times tan (8.6) - \dfrac{9.8\times 150^2}{2u^2cos^2(8.6)}$

$\dfrac{9.8\times 150^2}{2u^2cos^2(8.6)} = 10.29$

$\dfrac{9.8\times 150^2}{2\times 10.29\times cos^2(8.6)} = u^2$

$u = \sqrt{10959.34}$

u = 104.68 m/s

The initial speed of the ball is u = 104.68 m/s

8 0

3 years ago

Which is true of the buoyant force?

victus00 [196]

It acts in the upward direction.

8 0

3 years ago

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0

3 years ago

A train, traveling at a constant speed of 25.8 m/s, comes to an incline with a constant slope. While going up the incline, the t

zhannawk [14.2K]

The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

Answer:

Explanation:

So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.

So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then

v = 25.8 + (-1.66×8.3)

v =12.022 m/s.

So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

8 0

3 years ago