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Gala2k [10]
3 years ago
9

Question: 23 of 66: If you increase the frequency of a sound wave four times, what will happen to its speed?Select one of the op

tions below as your answer:. A. The speed will increase four times.. B. The speed will decrease four times.. C. The speed will remain the same.. D. The speed will increase twice.. E. The speed will decrease twice.
Physics
2 answers:
Marta_Voda [28]3 years ago
8 0

Answer:

C. The speed will remain the same.

Explanation:

As we know that speed of sound wave is given as

v = \sqrt{\frac{\gamma P}{\rho}}

now this speed of sound depends on the properties of medium like

\gamma = adiabatic constant of medium

P = pressure of medium

\rho = density of medium

so here we can say that sound speed is totally independent of frequency of sound so if we change the frequency of sound then the speed of sound will remain the same. so correct answer will be

C. The speed will remain the same.

zysi [14]3 years ago
4 0
If you increase the frequency of a sound wave four times, t<span>he speed will increase four times. The correct option among all the options that are given in the question is the first option or option "A". This also shows the frequency and speed of the waves are directly proportional to each other. I hope it helps you.</span>
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Volcanoes tend to erupt at places where
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it is 3

Explanation:

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<span>If we increase the force applied to an object and all other factors remain the same that amount of work will

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2 years ago
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Contrary to popular belief, a ski jumper does not achieve a large amount of "air" when doing a jump (less than 6 feet). This is
Elina [12.6K]

Answer:

The vertical distance that the ski jumper fell is 417.45 m.

Explanation:

Given;

initial horizontal velocity of the jumper, V_x = 26 m/s

horizontal distance of the jumper, dx = 240 m

The time of the motion is given by;

dx = Vₓt

t = dx / Vₓ

t = 240 / 26

t = 9.23 s

The vertical distance traveled by the diver is given by;

d_y = V_yt + \frac{1}{2}gt^2

initial vertical velocity, V_y, = 0

d_y =  \frac{1}{2}gt^2\\\\d_y = \frac{1}{2}(9.8)(9.23)^2\\\\d_y = 417.45 \ m

Therefore, the vertical distance that the ski jumper fell is 417.45 m.

6 0
3 years ago
a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?
julsineya [31]

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

  • u is the initial velocity of the object,
  • v is the final velocity of the object.
  • a is its acceleration, and
  • x is its displacement.

In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

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