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3 years ago

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Work done= ________ transferred

1 answer:

mamaluj [8]3 years ago

4 0

Answer:

Work done= Energy transferred

Explanation:

Work is the transfer of energy. In physics we say that work is done on an object when you transfer energy to that object. If you put energy into an object, then you do work on that object (mass).

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Some coastal regions of the world have cooler summers and warmer winters than inland regions at the same latitude. What accounts

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Water holds in heat very well. Keep the temperature more steady and average. The areas around the water will also have a less variant change in temperature as a result. This property of water is known as high specific heat.

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In a nuclear power plant, nuclear energy is first changed to ? energy

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Thermal energy Thermal energy <span>Thermal energy</span>

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3 years ago

At a height of ten meters above the surface of a freshwater lake, a sound pulse is generated. The echo from the bottom of the la

konstantin123 [22]

Answer:

Explanation:

Velocity of sound in air at 20 degree = 343 m/s

Velocity of sound in water at 20 degree = 1470 m/s

Time taken in to and fro movement in air

=( 2 x 10) / 343 = 0.0583 s

Rest of the time is

.171 - .0583 = .1127 s

This time is taken to cover distance in water. If d be the depth of lake

2d / velocity = time taken

2 d / 1470 = .1127

d = 82.83 m

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3 years ago

Assume you are in the car and the car is moving at a certain speed to

Setler79 [48]

You are at rest with respect to the car.
You are in motion with respect to the School.

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3 years ago

the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during

Ivanshal [37]

Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

Mass of the baseball after struck by the bat, M = 900 g

Velocity of the baseball after struck by the bat, v = 47 m/s

According to the conservation of momentum,

$Mv+mu=Mv_1+mv_2$

(900 x 47) + (200 x -30) = (900 x $v_1$ ) + (200 x $v_2$ )

36300 = (900 x $v_1$ ) + (200 x $v_2$ )

$9v_1 + 2v_2 = 363$ ..............(i)

$9v_1 = 363 - 2v_2$

$v_1=\frac{363 - 2v_2}{9}$

The mathematical expression for the conservation of kinetic energy is

$\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2$

$\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2$ ................(ii)

$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$

$21681 = 9v_1^2+2v_2^2$

Substituting (i) in (ii)

$21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2$

$(363-2v_2)^2+18v_2^2=195129$

$(363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0$

$22v_2^2-145v_2-63360=0$

Solving the equation, we get

$v_2=96 \ m/s, -30 \ m/s$

The negative velocity is neglected.

Therefore, substituting 96 m/s for $v_2$ in (i), we get

$v_1=\frac{363-(2 \times 96)}{9}$

= 19

Thus, only impulse of importance is used to find final velocity.

8 0

3 years ago