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horsena [70]
4 years ago
15

What is the relationship between mass and gravitational force between any two objects?

Physics
1 answer:
mamaluj [8]4 years ago
7 0
If two objects' mass increases, gravitational force pulls them down to the center more strongly because the greater the mass, the stronger the attraction is from the gravitational force from any two objects. An increase in then Earth's mass and increase in a man's mass would make gravitational force pull more strongly on the person because the Earth's gravitational force multiplied from the mass change and the man's greater mass pulls them down more quickly.
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What is a Super Massive Black Hole?
Serggg [28]

Answer:

I dont really know much but i know that it swallow anything it comes across in space.

4 0
3 years ago
Using the formula for calculating Pascal’s Principle, solve the following. Pressure(Pascal)=force(Newton)/area(meter2)
Verdich [7]
P= 5/30—-> P= .16667

Does that help?
8 0
4 years ago
An irrigation channel has a rectangular cross section of 1.5 ft deep x 11 ft wide on the input side. On the far end of the chann
garik1379 [7]

Answer:

55 ft/s

Explanation:

A₁ = Area of rectangular cross-section at input side = 1.5 x 11 = 16.5 ft²

A₂ = Area of rectangular cross-section at far end = 1.5 x 6 = 9 ft²

v₁ = speed of water at the input side of channel = 30 ft/s

v₂ = speed of water at the input side of channel = ?

Using equation of continuity

A₁ v₁ = A₂ v₂

(16.5) (30) = (9) v₂

v₂ = 55 ft/s

5 0
4 years ago
What is the moment of inertia of a cube with mass M=0.500kg and side lengths s=0.030m about an axis which is both normal (perpen
prisoha [69]

Answer:

The moment of inertia I is

I = 2.205x10^-4 kg/m^2

Explanation:

Given mass m = 0.5 kg

And side lenght = 0.03 m

Moment of inertia I = mass x radius of rotation squared

I = mr^2

In this case, the radius of rotation is about an axis which is both normal (perpendicular) to and through the center of a face of the cube.

Calculating from the dimensions of the the box as shown in the image below, the radius of rotation r = 0.021 m

Therefore,

I = 0.5 x 0.021^2 = 2.205x10^-4 kg/m^2

8 0
4 years ago
9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o
Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).

Now the force of friction is F_s=\mu*N, where N is the normal force and from the diagram it is F_y=mg*cos(\theta).

Thus F_s=\mu*N=\mu*mg*cos(\theta).

Therefore,

F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).

Substituting the value for F_H,m,\mu, and \:\theta we get:

F_{tot}= -38.63N.

Now acceleration is simply

a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.

The negative sign indicates that the acceleration is directed up the incline.

Part B:

d=\frac{1}{2} at^2

Which can be rearranged to solve for t:

t=\sqrt{\frac{2d}{a} }

Substitute the value of d=0.50m and a=7.7m/s and we get:

t=0.36s.

which is our answer.

Notice that in using the formula to calculate time we used the positive value of a, because for this formula absolute value is needed.

5 0
4 years ago
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