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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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Any energy transformation involves the loss of some energy as B. Heat.
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3 years ago
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A ball has a mass of 2 kg and is thrown with a force of 8 Newtons for .35 seconds. What is the ball's change in
Anna [14]

Answer:

1.4s

Explanation:

Given parameters:

Mass of ball  = 2kg

Force  = 8N

Time  = 0.35s

Unknown:

Change in velocity  = ?

Solution:

To solve this problem, we use the expression obtained from Newton's second law of motion which is shown below:

      Ft  = m(v  - u)

 So;

         Ft  = m Δv

F is the force

t is the time

m is the mass

Δv is the change in velocity

             8 x 0.35  = 2 x Δv

                  Δv  = 1.4s

6 0
2 years ago
In Fig. 4-41, a ball is thrown up onto a roof, landing 4.00 s later at height h ???? 20.0 m above the release level. The ball’s
Yanka [14]

"Fig is attacted with answer"

Answer:

a) d = 33.72 m

b) v_{i} = 26 m/s

c) β = 71.08°

Explanation:

a)

When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.

Given data:

time = t = 4.00 s

Height = h = 20 m

Angle = θ = 60°

Horizontal distance = d = ?

Using 2nd  equation of motion

h = v_{y_{f}}t + \frac{1}{2}gt^{2}

-20 = v_{y_{f}} (4) + 0.5(-9.8)(4)²

v_{y_{f}} (4) = 58.4

v_{y_{f}}  = 14.6 m/s

This is vertical component of velocity when the ball is on the roof. To calculate the Final velocity and horizontal component, we use

v_{f} = v_{y_{f}} / sinθ

v_{f} = 14.6 / sin 60

v_{f} = 16.86 m/s

v_{x_{f}} = v_{f}cosθ

v_{x_{f}} = 16.86 cos 60

v_{x_{f}} = 8.43 m/s

To calculate the horizontal distance

d = v_{y_{f}} t

d = (8.43)(4)

d = 33.72 m

b)

We know the values of Landing angle, height of roof, time of flight. In part a, We calculate the landing velocity of the ball and also its horizontal and vertical component. As the ball followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

So,

v_{x_{f}} = v_{x_{i}}

but the vertical component of velocity vary with and there is an acceleration along vertical direction which is equal to gravitation acceleration g.

So,

g = (v_{y_{f}} - v_{y_{i}} ) / t

9.8 =  14.6 - v_{y_{i}}) / 4

v_{y_{i}} = 24.6 m/s

v_{i} = \sqrt{v_{x_{i}}^{2}+v_{y_{i}}^{2} }

v_{i} = \sqrt{8.43^{2}+24.6^{2}}

v_{i} = 26 m/s

c)

cos β = v_{x_{i}} / v_{i}

β = cos⁻¹ (8.43 / 26)

β = 71.08°

3 0
2 years ago
What happens to the equilibrium price and quantity for jelly when the price of peanut butter increases? (Assume that peanut butt
Darya [45]

Answer:

Equilibrium price will fall and quantity demanded for jelly will also fall

<u>Explanation:</u>

Peanut butter and jelly are consumed together. It means they are complementary goods. These are those goods which are bought together. So increase or decrease in the price of one commodity will automatically affect the demand for another commodity.

When the price of peanut butter increases than people will demand less of peanut butter. Similarly, the demand for jelly is associated directly with the demand for peanut butter. So it will also fall . Due to the fall in the price of jelly and simultaneous fall in demand, the equilibrium price will fall.

5 0
3 years ago
the sole of a tennis shoe has a surface area of 0.0290 m^2. if it is worn by a 65.0 kg person, what pressure does the shoe exert
AURORKA [14]

Answer: 21965.517 Pa

Explanation:

Pressure P is the force F exerted by a gas, a liquid or a solid on a surface (or area) A, its unit is Pascal Pa which is equal to N/m^{2} and its formula is:  

P=\frac{F}{A} (1)

In this case we have the surface of a sole of a tennis shoe:

A=0.0290 m^{2} (2)

And the mass m of the person who wears it:

m=65 kg

On the other hand, we know the weight is the force  F the Earth exerts on people and objects due gravity g :

F=m.g=(65 kg)(9.8m/^{2})

F=637N (3)

Substituting (2) and (3) in (1):

P=\frac{637N}{0.0290 m^{2}} (4)

Finally:

P=21965.517 Pa This is the pressure the shoe exert on the ground

5 0
3 years ago
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