When a photon hits an electron <span>the photon will be partially </span>absorbed<span>and the electron emits another photon with lower energy.</span>
Answer:
<em>The atoms in the hot bodies have higher kinetic energy than those of the cold bodies. Thus to maintain thermal equilibrium, the atoms of higher kinetic energy tries to move and collide with the atoms of low kinetic energy. Thus heat transfers from a hot body to a cold body.</em>
<em>Explanation:</em>
<em>Explanation:The atoms in the hot bodies have higher kinetic energy than those of the cold bodies. Thus to maintain thermaler kinetic energy tries to move and collide with the atoms of low kinetic energy. Thus heat transfers from a hot body to a cold body. </em>
Answer:
Between 2.0 s and 4.0 s (B and C)
Between 5.0 s and 8.0 s (D and E)
Between 10.0 s and 11.0 s (F and G)
Explanation:
The graph shown in the figure is a velocity-time graph, which means that:
- On the x-axis, the time is plotted
- On the y-axis, the velocity is plotted
Therefore, this means that the object is not moving when the line is horizontal (because at that moment, the velocity is constant, so the object is not moving). This occurs in the following intervals:
Between 2.0 s and 4.0 s (B and C)
Between 5.0 s and 8.0 s (D and E)
Between 10.0 s and 11.0 s (F and G)
From the graph, it would be possible to infer additional information. In particular:
- The area under the graph represents the total distance covered by the object
- The slope of the graph represents the acceleration of the object
Put a fork under your pillow tonight, and your wish will come true tomorrow.
Answer:
W = 439998 J = 439.99 KJ
Explanation:
First, we will calculate the acceleration of the car by using the first equation of motion:
![v_f = v_i + at\\\\a = \frac{v_f-v_i}{t}](https://tex.z-dn.net/?f=v_f%20%3D%20v_i%20%2B%20at%5C%5C%5C%5Ca%20%3D%20%5Cfrac%7Bv_f-v_i%7D%7Bt%7D)
where,
a = acceleration = ?
vf = final speed =
= 30 m/s
vi = initial speed =
= 10 m/s
t = time = 30 s
Therefore,
![a = \frac{30\ m/s - 10\ m/s}{30\ s}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B30%5C%20m%2Fs%20-%2010%5C%20m%2Fs%7D%7B30%5C%20s%7D)
a = 0.67 m/s²
Now, we will calculate the force applied by the engine:
F = ma
where,
F = force = ?
m = mass = 1100 kg
Therefore,
F = (1100 kg)(0.67 m/s²)
F = 733.3 N
Now, we will calculate the distance covered by the car by using the second equation of motion:
![s = v_it+\frac{1}{2}at^2\\\\s = (10\ m/s)(30\ s)+\frac{1}{2} (0.67\ m/s^2)(30\ s)^2](https://tex.z-dn.net/?f=s%20%3D%20v_it%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5C%5Cs%20%3D%20%2810%5C%20m%2Fs%29%2830%5C%20s%29%2B%5Cfrac%7B1%7D%7B2%7D%20%280.67%5C%20m%2Fs%5E2%29%2830%5C%20s%29%5E2)
s = 600 m
Now, the work done (W) by engine can be calculated as follows:
W = Fs
W = (733.3 N)(600 m)
<u>W = 439998 J = 439.99 KJ</u>