According to the electromagnetic spectrum,

Physics

2 answers:

olga_2 [115]4 years ago

5 0

A. long-wave radiation occupies a smaller percentage than short-wave radiation

dezoksy [38]4 years ago

4 0

From the Electromagnetic Spectrum:

Radio, Micro wave, Infrared, Visible light, Ultra Violet, X rays, Y rays.

Only the Radio wave of wavelength 10³ m, has a positive power index, others have a negative exponent.

So these long waves form a smaller percentage.

Option A is correct. Others are not true.

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Why do some athletes engage in cross training

tigry1 [53]

Answer; Cross training enables your body to recuperate faster from injuries, in some cases because other exercises can directly improve the condition caused by your regular activity. For example, Achilles tendonitis, caused by overuse, can be improved by eccentric strengthening of the calf muscles.Explanation:

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3 years ago

A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th

Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(4)}{2}(2(0)+4)$

$F = 25000 lb$

b) Force on deep end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(10)}{2} (2(0)+10)$

$F = 187500 lb$

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

$F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)$