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Elanso [62]
3 years ago
6

The diagram shows a wave. The quantity shown is the amplitude. This quantity is directly proportional to which of these other qu

antities? A) Energy B) Frequency C) Period D) Wavelength
Physics
2 answers:
Hoochie [10]3 years ago
4 0

Answer:

A. Energy

Explanation:

Katen [24]3 years ago
3 0

As we know that frequency and wavelength are dependent on each other

and this is given by

f = \frac{c}{\lambda}

here we know that

\lambda = wavelength

f = frequency

c = speed of wave

while for energy and intensity of wave we know that

Energy = KA^2

here A = amplitude of wave

so energy directly depends on amplitude of wave

so correct answer will be

<em>A) Energy</em>

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MrMuchimi

Answer:

The answer is ball D

Explanation:

because I said it was

8 0
4 years ago
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A boy sleds down a hill and onto a frictionless ice- covered lake at 10.0 m/s. In the middle of the lake is a 1000-kg boulder. W
mina [271]

Answer:

The speed of the sled is 9.2 m/s

The speed of the boulder is 0.82 m/s

Solution:

As per the question:

Mass of the boulder, m_{B} = 1000\ kg

Mass of the sled, m_{S} = 2.50\ kg

Mass of the boy, m_{b} = 40\ kg

Initial Velocity, v = 10.0 m/s

Now,

To calculate the speed of both the sled and the boulder after the occurrence of the collision:

m = m_{b} + m_{S} = 40 + 2.50 = 42.50\ kg

Initial velocity of the boulder, v_{B} = 0\ m/s

Since, the collision is elastic, both the energy and momentum rem,ain conserved.

Now,

Using the conservation of momentum:

mv + m_{B}v_{B} = mv' + m_{B}v'_{B}

where

v' = final velocity of the the system of boy and sled

v'_{B} = final velocity of the boulder

42.50\times 10 + m_{B}.0 = 42.50v' + 1000v'_{B}

42.50v' + 1000v'_{B} = 425            (1)

Now,

Using conservation of energy:

\frac{1}{2}mv^{2} + \frac{1}{2}m_{B}v_{B}^{2} = \frac{1}{2}mv'^{2} + \frac{1}{2}m_{B}v'_{B}^{2}

42.50\times 10^{2} + m_{B}.0 = 42.50v'^{2} + 1000v'_{B}^{2}

42.50v'^{2} + 1000v'_{B}^{2} = 4250         (2)

Now, from  eqn (1) and (2):

v' = \frac{m - m_{B}}{m + m_{B}}\times v

v' = \frac{42.50 - 1000}{42.5 + 1000}\times 10 = - 9.2\ m/s

Now,

v'_{B} = \frac{2m}{m + m_{B}}\times v

v'_{B} = \frac{2\times 42.50}{42.5 + 1000}\times 10 = 0.82\ m/s

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H2 is the correct answer
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