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3 years ago

Which factors affect gravitational force? Check all that apply.

Physics

2 answers:

alina1380 [7]3 years ago

6 0

Answer:

distance between the objects

masses of the objects

Explanation:

Gravitational force is a force of attraction that pulls two objects with masses together.

To best understand the concept of gravitational force, newton's law of universal gravitation provides a good insight. The law states that "every object in the universe attracts each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distances between them".

From the law, we see that gravitational force is dependent on the masses of the object and their distances.

sertanlavr [38]3 years ago

4 0

Answer:

B. distance between the objects

C. masses of the objects

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collision occurs betweena 2 kg particle traveling with velocity and a 4 kg particle traveling with velocity. what is the magnitu

anzhelika [568]

Answer:

metre per seconds

Explanation:

because velocity = distance ÷ time

4 0

3 years ago

After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s. T

motikmotik

Answer:

V' = 0.84 m/s

Explanation:

given,

Linear speed of the ball, v = 2.85 m/s

rise of the ball, h = 0.53 m

Linear speed of the ball, v' = ?

rotation kinetic energy of the ball

$KE_r = \dfrac{1}{2}I\omega^2$

I of the moment of inertia of the sphere

$I = \dfrac{2}{5}MR^2$

v = R ω

using conservation of energy

$KE_r = \dfrac{1}{2}( \dfrac{2}{5}MR^2)(\dfrac{V}{R})2$

$KE_r = \dfrac{1}{5}MV^2$

Applying conservation of energy

Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy

$\dfrac{1}{2}MV^2 + \dfrac{1}{5}MV^2 = \dfrac{1}{2}MV'^2 + \dfrac{1}{5}MV'^2 + M g h$

$0.7 V^2 = 0.7 V'^2 + gh$

$0.7\times 2.85^2 = 0.7\times V'^2 +9.8\times 0.53$

V'² = 0.7025

V' = 0.84 m/s

the linear speed of the ball at the top of ramp is equal to 0.84 m/s

6 0

3 years ago

The forces in (Figure 1) are acting on a 1.0 kg object.What is ax , the x -component of the object's acceleration

a_sh-v [17]

The x -component of the object's acceleration is 2 m/s².

<h3>What's the resultant force along x- direction?</h3>

Forces along x axis direction are as follows

4N along +x axis, so it's taken as +4 N
2N along -x axis , so it's taken as -2N.

Resultant force along x direction = 4N - 2N = 2 N which is along + ve x direction.

<h3>What's the acceleration along x axis direction?</h3>

As per Newton's second law, Force = mass × acceleration of the object
Force along x axis= mass × acceleration along x axis= 2N
Acceleration = 2/ mass = 2/1 = 2 m/s²

Thus, we can conclude that the acceleration along x axis is 2 m/s².

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?

Learn more about the acceleration here:

brainly.com/question/460763

#SPJ1

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2 years ago

The speed of the center of the earth as it orbits the sun is 107257 kmph and the absolute angular velocity of the earth about it

Sindrei [870]

Answer with Explanation:

We are given that

Speed , $v_0$ =107257 kmph

Angular velocity, $\omega=7.292\times 10^{-5} rad/s$

Radius of earth,r= $6371 km=6371000 m$

1 km=1000 m

Linear velocity, $v=r\omega=6371000\times 7.292\times 10^{-5}=464.57 m/s$

Linear velocity, $v=464.57\times \frac{18}{5}=1672.46 km/h$

Velocity at point A, $v_A=-vi+v_0 j=-1672.46 i+107257j kmph$

Velocity at point B, $v_B=v_0j-vj=107257j-1672.46j=105584.54j kmph$

Velocity at point C, $v_C=v_0j+vi=1672.46 i+107257j kmph$

Velocity at point D, $v_D=v_0j+vj=107257j+1672.46j=108929.46jkmph$

3 0

3 years ago

The overhang beam is subjected to the uniform distributed load having an intensity of w = 50 kn/m. determine the maximum shear s

alex41 [277]

Given:

Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.

We need to determine the maximum shear stress developed in the beam:

τ = F/A

Assuming the area of the beam is 100 m^2 with a length of 10 m.

τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2<span />

8 0

4 years ago