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If two objects have the same mass, what determines the strength of the gravitational force between them? (4 points) Select one:

Evgen [1.6K]

According to this equation

F = G × m₁*m₂ ÷ r²

other than the mass, the distance also affects the gravitational force between two objects (same mass or not).

Therefore the correct answer is B. The distance between the objects

Future note* use formulas to help you figure these sort of questions out. (if they have a formula to begin with).

3 0

2 years ago

A ball dropped from a bridge takes three seconds to reach the water below how far is the bridge above the water?

tatuchka [14]

Given that:

Ball dropped from a bridge at the rate of 3 seconds

Determine the height of fall (S) = ?

As we know that, S = ut + 1/2 ×a.t²

u =initial velocity = 0

a= g =9.81 m/s (since free fall)

S = 0+ 1/2 × 9.81 × 3²

S = 44.145 m

44.145 m far is the bridge from water

6 0

2 years ago

Graphs help you see ______.

Anastaziya [24]

D- if i was right please mark me as a brainliest answer and a thank you

4 0

3 years ago

Read 2 more answers

Changes in behavior of managers, coworkers, and subordinates can be a warning sign before a layoff. True or false

Yakvenalex [24]

Answer:

The correct answer is - true.

Explanation:

Changes in the higher authorities or managers or coworkers can be a warning sign before a layoff, however, a company or organization needs to give an official notice before a particular time period before the layoff.

The layoff is a temporary suspension or permanent termination from the employment of an employee due to reasons related to business such as reducing manpower or staff and less work in the organization. These reasons can lead to a change in the behavior of the managers, coworkers, and subordinates.

3 0

2 years ago

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Fantom [35]

a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

$W=mg$

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

$\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta$

where L is the length of the pencil.

The relationship between torque and angular acceleration $\alpha$ is

$\tau = I \alpha$ (1)

where

$I=\frac{1}{3}mL^2$

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0

2 years ago