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34kurt
3 years ago
7

How are electromagnetic waves like ocean waves

Physics
1 answer:
Strike441 [17]3 years ago
8 0

Answer:

Another way electromagnetic (EM) waves are like ocean waves is that they both move energy from one place to another. The biggest difference between ocean waves and EM waves is that ocean waves move across water. Energy moves through the water. EM waves don't need anything to move through.

Explanation:

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A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra
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Answer:

V_{ft}= 317 cm/s

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

P_i = P_f

Where:

P_i=M_cV_{ic} + M_tV_{it}

P_f = M_cV_{fc} + M_tV_{ft}

Now:

M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}

Where M_c is the mass of the car, V_{ic} is the initial velocity of the car, M_t is the mass of train, V_{fc} is the final velocity of the car and V_{ft} is the final velocity of the train.

Replacing data:

(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}

Solving for V_{ft}:

V_{ft}= 3.17 m/s

Changed to cm/s, we get:

V_{ft}= 3.17*100 = 317 cm/s

b) The kinetic energy K is calculated as:

K = \frac{1}{2}MV^2

where M is the mass and V is the velocity.

So, the initial K is:

K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2

K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2

K_i = 22.06 J

And the final K is:

K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = 19.61 J

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

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