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n200080 [17]
3 years ago
8

1) Arrange the following spectral regions in order of increasing energy: infrared, microwave, ultraviolet, visible.

Physics
1 answer:
Anton [14]3 years ago
7 0

Answer:

The order of increasing energy is as follows

"microwave < infrared < visible < ultraviolet"

Option (A) is correct.

Explanation:

Given:

Arrange the following spectral regions in order of increasing energy: infrared, microwave, ultraviolet, visible.

From the formula of energy in terms of frequency.

  E = hf

Where h = planck constant, f = frequency of light.

From above formula we can conclude that higher frequency means higher energy.

In our case ultraviolet has higher frequency and microwave has lower frequency.

So ultraviolet has higher energy and microwave has lower energy.

microwave < infrared < visible < ultraviolet

Therefore, the order of increasing energy is as follows

"microwave < infrared < visible < ultraviolet"

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The cable of the 1800kg elevator cab in Fig. 8−56 snaps when the cab is at rest at the first floor, where the cab bottom is a di
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a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

a ) The speed of the cab just before it hits the spring,

Ki + Pi = K final + P final + W

1 / 2 m vo² + m g hi = 1 / 2 m v² + m g h + f d

0 + ( 1800 * 9.8 * 3.7 m ) = ( 1 / 2 * 1800 * v² ) + 0 + ( 4400 * 3.7 )

v = 7.4 m / s

b ) The maximum distance x that the spring is compressed,

Ki + Pi = K final + P final + W + Fs

1 / 2 m vo² + m g x = 1 / 2 m v² + m g h + f d + 1 /2 k x²

( 1/2 * 1800 * 7.4² ) + ( 1800 * 9.8 * x ) =0 + 0+ ( 4400 * x ) + ( 1/2 * 1800 * x² )

75000 x² + 13420 x - 50625 = 0

x = 0.9 m

c ) The distance that the cab will bounce back up the shaft,

Ki + Pi + Fs = K final + P final + W

1 / 2 m vo² + m g x + 1 /2 k x² = 1 / 2 m v² + m g h + f d

0 + 0 + ( 1 / 2 * 0.15 * 0.9² ) = 0 + ( 1800 * 9.8 * h ) + ( 4400 * h )

h = 2.8 m

Therefore,

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

To know more about law of conservation of energy

brainly.com/question/12050604

#SPJ4

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