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3 years ago

What is happening in the graph shown below?

Physics

1 answer:

Licemer1 [7]3 years ago

7 0

Answer:

D. The object moves away from the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 2 m/s.

Explanation:

I just got it right lol

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A permanent magnet is placed near a bar of soft iron PQ what will happen

boyakko [2]

Answer:

it will rust

Explanation:

hope it helps

4 0

3 years ago

Use your understanding of heat loss to ESTIMATE the cost of the lost energy through one standard window during an average summer

miskamm [114]

Answer:

The cost of energy is $ 0.34.

Explanation:

The energy is the capacity to do work.

The energy is a scalar quantity and its SI unit is Joule.

The commercial unit of energy is kWh.

Cost of 1 kWh energy = $ 0.17

energy loss by standard window is 2 kWh .

So, the cost of lost of energy is

Cost = $ 0.17 x 2 = $ 0.34

3 0

3 years ago

Your friend wants to join the school track team, and has asked for your help to determine how fast she can run. 4. What kind of

DENIUS [597]

Answer:

I think you would need too time your friend and know how too divide that time by other racers to see how much faster she needs to get. I THINK.

7 0

2 years ago

A 0.150-kg lump of clay is dropped from a height of 1.15 m onto the floor. it sticks to the floor and does not bounce.

Serggg [28]

Missing question: "<span>What is the magnitude of the impulse I imparted to the clay by the floor during the impact?"

Solution:
The impulse is equal to the variation of momentum of the object:
</span> $I=\Delta p=m \Delta v$
<span>where m is the mass of the object and </span> $\Delta v$ is the variation of velocity of the object during the impact.

By using energy conservation, we can calculate the velocity of the object before the impact. In fact, the initial potential energy of the object is all converted into kinetic energy before the impact:
$mgh= \frac{1}{2}mv^2$
from which we find v:
$v= \sqrt{2gh}= \sqrt{2 \cdot 9.81 m/s^2 \cdot 1.15 m}=4.75 m/s$

This is the velocity of the object just before the impact. After the impact, the object comes to rest: this means that the variation of velocity of the object is equal to 4.75 m/s. Therefore, the impulse is

$I=m \Delta v=(0.150 kg)(4.75 m/s)=0.71 Kg m s^{-1}$

6 0

3 years ago

What can we say about the forces? A. They are balanced. B. They are unbalanced. C. They are not in equilibrium.

yulyashka [42]

You can say anything you want to about them without fear of contradiction. There's no information here to prove you wrong.

5 0

3 years ago