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2 years ago

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A 1200 kg car is moving at 8 m/s when is strikes a barrier with an equivalent spring constant of 18,000 N/m. How far is the barr

ier compressed?

1 answer:

kenny6666 [7]2 years ago

8 0

1200 ➗8

<h2> that gives 150</h2>

i think i am not correct

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Rainfall occurs on the earth due to effect of gravity

Anna [14]

<em>Answer:</em>

<h3><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>True</em></h3>

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2 years ago

2 ways to change frictional force between 2 objects

Zanzabum

In order to change the frictional force between two solid surfaces, it can be changed by shorter distances and by the amount of weight it has or the amount of force that is pushing that object to go however distance it can.

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3 years ago

Read 2 more answers

Why would north america will not be able to view the eclipse

Flauer [41]

The North America will not be able to view the eclipse because of its location on the earth caused by the tilting of the earth.

<h3>Effect of tilting of the Earth</h3>

The tilt of the Earth is what causes seasons to occur. These are the seasons in relation to the Northern Hemisphere.

The tilting of the Earth causes the difference in the amount of sun reaching each region of the earth like in the North America continent.

Thus, the North America will not be able to view the eclipse because of its location on the earth caused by the tilting of the earth.

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2 years ago

Suppose a constant net force of 345 N is applied to slide a heavy stationary couch across the

ira [324]

Answer:

517.5Ns

Explanation:

F=(MV - MU)/t

where MV - MU is the change in momentum,

therefore, MV - MU = Ft

= 345 X 1.

= 517.5Ns

4 0

2 years ago

A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.

denpristay [2]

Answer:

Approximately $5.5\; \rm N$ , assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let $q_1$ and $q_2$ denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, $q_1 = 20 \times 10^{-6}\; \rm C$ and $q_2 = 15 \times 10^{-6}\; \rm C$ .

Let $r$ denote the distance between these two point charges. In this question, $r = 0.7\; \rm m$ .

Let $k$ denote the Coulomb constant. In standard units, $k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}$ .

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}$ .

Substitute in the values and evaluate:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}$ .

8 0

3 years ago