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Svetach [21]
3 years ago
7

How the behavior of waves is affected by a medium

Physics
1 answer:
Schach [20]3 years ago
3 0
The medium determines the speed of the wave traveling in it, which also can have a number of other effects, including how much the wave bends (refracts), whether it reflects, etc.
Because waves move through space, they must have a velocity. The velocity of a wave is a function of the type of wave, and the medium it travels through. Electromagnetic waves moving through a vacuum, for instance, travel at roughly 3 x
10
8
m/s. This value is so famous and common in physics it is given its own symbol, c.
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) 5 -5 1 2 3 4 5 Other than at t = 0, when is the velocity of the object equal to zero? 1. 5.0 s 2. 4.0 s 3. 3.5 s 4. At no other time on this graph. correct 5. During the interval from 1.0 s to 3.0 s. Explanation: Since vt = Z t 0 a dt, vt is the area between the acceleration curve and the t axis during the time period from 0 to t. If the area is above the horizontal axis, it is positive; otherwise, it is negative. In order for the velocity to be zero at any given time t, there would have to be equal amounts of positive and negative area between 0 and t. According to the graph, this condition is never satisfied. 005 (part 1 of 1) 0 points Identify all of those graphs that represent motion at constant speed (note the axes carefully). a) t x b) t v c) t a d) t v e) t a 
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3 years ago
A hiker is at the bottom of a canyon facing the canyon wall closest to her. She is 790.5 meters from the wall and the sound of h
tester [92]

Answer:

4.80 seconds

Explanation:

The velocity of sound is obtained from;

V= 2d/t

Where;

V= velocity of sound = 329.2 ms-1

d= distance from the wall = 790.5 m

t= time = the unknown

t= 2d/V

t= 2 × 790.5/ 329.2

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7 0
3 years ago
A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9
Andreyy89

Answer:

d = 0.0306 m

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

U + K = constant

\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0

7.05 + 30.5 = \frac{0.266}{r}

r = 7.08 \times 10^{-3} m

so distance moved by the particle is given as

d = r_1 - r_2

d = 0.0377 - 0.00708

d = 0.0306 m

6 0
3 years ago
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