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Pie
3 years ago
10

A student went to a hill station early in the morning, he could hear the echo of his clap after 0.1 second. When he went to the

same place in the afternoon, he could not hear the echo at all, explain the reason for his changed behaviour.
Physics
1 answer:
Dahasolnce [82]3 years ago
7 0

Answer:

See the explanation.

Explanation:

The speeds of sound depend upon the temperature of the medium. As the temperature increased in the afternoon, the speed of sound increases. Then the time taken by reflected sound will be less than 0.1 sec in the afternoon. And to hear an echo the time gap between an original sound and reflected sound must be at least 0.1 sec. That is why the student could not hear the echo at all in the afternoon.

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The law of conversation of energy states that energy cannot be created or
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Answer:

It says energy can't be created or destroyed

Explanation:

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Ice that formed thousands of years ago is often found to contain tiny bubbles of gas. this gas came from __________. ice that fo
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A vertical spring with a spring constant of 420 N/m is mounted on the floor. From directly above the spring, which is unstrained
mixas84 [53]

Answer:

19.53 cm

Explanation:

The computation of the height is as follows:

Here we applied the conservation of the energy formula

As we know that

P.E of the block = P.E of the spring

 m g h = ( 1 ÷ 2) k x^2

where

m = 0.15

g = 9.81

k = 420

x = 0.037

So now put the values to the above formula

(0.15) (9.81) (h) = 1 ÷2 × 420 × (0.037)^2

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3 years ago
Define orbital velocity
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8 0
3 years ago
A spring stretches by 0.0208 m when a 3.39-kg object is suspended from its end. How much mass should be attached to this spring
sashaice [31]

Answer:

COMPLETE QUESTION

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?

Explanation:

Given that,

Extension of spring

x = 0.0208m

Mass attached m = 3.39kg

Additional mass to have a frequency f

Let the additional mass be m

Using Hooke's law

F= kx

Where F = W = mg = 3.39 ×9.81

F = 33.26N

Then,

F = kx

k = F/x

k = 33.26/0.0208

k = 1598.84 N/m

The frequency is given as

f = ½π√k/m

Make m subject of formula

f² = ¼π² •(k/m

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Then, m4π²f² = k

So, m = k/(4π²f²)

So, this is the general formula,

Then let use the frequency above

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4 0
3 years ago
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