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3 years ago

5

What name is given to the property of an object to resist changes in motion

1 answer:

Rasek [7]3 years ago

3 0

The anwser is newtons 1st law

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What do objects in space that are moving at a constant velocity in a straight line do?

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Answer:

B

Explanation:

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3 years ago

An object of weight W is in freefall close to the surface of Earth. The magnitude of the force that the object exerts on Earth i

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Earth is B)equal to W

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2 years ago

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1. Car A
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2 years ago

An underwater scuba diver sees the Sun at an apparent angle of 43.0° above the horizontal. What is the actual elevation angle of

Inessa05 [86]

Answer:

The actual elevation angle is 12.87 degrees

Explanation:

In the attachment you can clearly see the situation. The angle of elevation as seen for the scuba diver is shown in magenta, we conclude that $\theta_2=90-43=47$ .

Using Snell's Law we can write:

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

$\implies \sin(\theta_1)=\frac{n_2}{n_1}\sin(\theta_2)$ ,

Let's approximate the index of refraction of the air (medium 1 in the picture) to 1.

We thus have:

$\sin(\theta_1)=n_2\sin(\theta_2)=1.333\sin(47)$

$\implies\theta_1=\arcsin[n_2\sin(\theta_2)]=\arcsin[1.333\sin(47)]\approx 77.13$ . Calling $\alpha$ the actual angle of elevation, we get from the picture that $\alpha=90-77.13=12.97$

7 0

2 years ago

A(n) 7.7-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting th

-BARSIC- [3]

Answer:

Average acceleration on first part of the chunk is given as

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = -13.125 m/s^2$

Explanation:

By momentum conservation along x direction we will have

$mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2$

so we have

$v_1 + v_2 = 2v$

$v_1 + v_2 = 4.68$

also by energy conservation

$\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J$

$\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17$

$(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)$

$(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83$

$21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83$

$2v_2^2 - 9.36v_2 + 2.12 = 0$

by solving above equation we will have

$v_1 = 4.44 m/s$

$v_2 = 0.24 m/s$

Average acceleration on first part of the chunk is given as

$a_1 = \frac{4.44 - 2.34}{0.16}$

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = \frac{0.24 - 2.34}{0.16}$

$a_2 = -13.125 m/s^2$

5 0

3 years ago