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2 years ago

A 10 kg brick and a 1 kg book are dropped in a vacuum. The force of gravity on the 10 kg brick is what?

2 answers:

7 0

10 times as much. Since F=m*a, and a is constant, the only thing that affects force is the mass.

In response to the below answer, the acceleration due to gravity does not change. The force due to gravity definitely DOES change depending on the mass of the object. Since the force is what the problem asks for, the answer is 10

LuckyWell [14K]2 years ago

3 0

If this is happening on Earth, then the force of gravity between the Earth and the brick is 98 Newtons. It doesn't matter whether the brick is falling, rising, dropped in a vacuum, in water, or into chicken soup, or what else is dropped near it. None of that has any effect on the gravitational force between the Earth and the brick.

You might be interested in

Name two ways to reduce friction and two ways to increase it.

Maurinko [17]

Ways to increase friction

- increase the roughness of the contact materials 
- increase the pressure on the contact 

Ways to decrease friction 

- float the moving body on air 
- suck out any air

4 0

2 years ago

In a lab, four balls have the same velocities but different masses.

olya-2409 [2.1K]

Answer:

New Momentum of Ball B $=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}$

Explanation:

Given:

Mass of Ball A=1kg

Mass of Ball B= 2kg

Mass of Ball C=5kg

Mass of Ball D=7kg

Velocities of A=B=C=D=2.2 $\frac{m}{s}$

Momentum of Ball A=2.2 $\frac{k g m}{s}$

Momentum of Ball B=4.4 $\frac{k g m}{s}$

Momentum of Ball C=11 $\frac{k g m}{s}$

Momentum of Ball D=15 $\frac{k g m}{s}$

To Find:

Change in Momentum When of Ball B gets tripled

Solution:

Though all balls have same velocity, thus we get

Velocities of A=B=C=D=2.2 $\frac{m}{s}$

Initial Momentum of Ball B=4.4 $\frac{k g m}{s}$

If the Mass of Ball B gets tripled;

We get New Mass of Ball B=3×Actual Mass of the ball

=3×2=6kg

Thus we get Mass of Ball B=6kg

According to the formula,

Change in momentum of Ball B $\Delta p=m \times \Delta v$

Where $\Delta p$ =change in momentum

m=mass of the ball B

$\Delta v$ =change in velocity ball B

And $\Delta v=v,$ since all balls, have same velocity

Thus the above equation, changes to

$\Delta p=m \times v$

Substitute all the values in the above equation we get

$\Delta p=6 \times 2.2$

$=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}$

Result:

Thus the New Momentum of ball B $=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}$

3 0

3 years ago

Read 2 more answers

. If she

I am Lyosha [343]

Answer:

9 meters

Explanation:

Given:

Mass of Avi is, $m=40\ kg$

Spring constant is, $k=176,400\ N/m$

Compression in the spring is, $x=20\ cm=0.20\ m$

Let the maximum height reached be 'h' m.

Now, as the spring is compressed, there is elastic potential energy stored in the spring. This elastic potential energy is transferred to Avi in the form of gravitational potential energy.

So, by law of conservation of energy, decrease in elastic potential energy is equal to increase in gravitational potential energy.

Decrease in elastic potential energy is given as:

$EPE=\frac{1}{2}kx^2\\EPE=\frac{1}{2}\times 176400\times (0.20)^2\\EPE=88200\times 0.04=3528\ J$