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2 years ago

9

Which three characteristics make a glowing piece of charcoal a blackbody radiator? A. The frequencies of EMR it emits depend on

its temperature. B. It emits only one frequency of EMR. C. It absorbs most of the EMR it receives. D. It emits a range of wavelengths of EMR.

1 answer:

Eva8 [605]2 years ago

4 0

Answer:

A. The frequencies of EMR it emits depend on its temperature

B. It emits only one frequency of EMR

C. It absorbs most of the EMR it receives

Explanation:

A blackbody is an object that absorbs most of the electromagnetic spectrum of energy that falls on it.

According to law to reradiates most of the available energy back on top the outer space at an efficiency of 100% and thus radiation may be in the visible range of temperature than are in 1000K.

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Glass absorbs ultraviolet (UV) rays from the Sun. Would a fraction of the incident UV light be reflected from the air/glass boun

fiasKO [112]

Answer:

Yes

Explanation:

Any transparent surface in practical is neither a perfect absorber of electromagnetic waves neither a perfect reflector. Generally all the transparent surfaces reflect some amount of irradiation and the other parts are absorbed and transmitted.

<u>That is given by as relation:</u>

$\alpha+\rho+\tau=1$

where:

$\alpha=$ absorptivity which is defined as the ratio of the absorbed radiation to the total irradiation

$\rho=$ reflectivity is defined as the ratio of reflected radiation to the total irradiation

$\tau=$ transmittivity is defined as the ratio of total transmitted radiation to the total irradiation

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3 years ago

Use your knowledge of waves to explain why echoes occur. Use your explanation to devise a system to measure distances to objects

german

Explanation:

Echoes occur due to the reflection of sound from any obstacle, but not all the reflected sound waves lead to the phenomenon of echo. For the echo to be heard it actually depends upon the human perception as well, human ears can encounter the difference between the sound wave directly form the source and the reflected sound waves only if there is a minimum time gap of one-tenth of a second. For this time gap in the atmosphere at normal temperature and pressure the obstacle must be at least 7 meters away from the sound source.

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2 years ago

Describe how the properties of sound waves change as they spread out in a spherical pattern.

guapka [62]

They go up and down

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2 years ago

The gravitational force between two objects (mass1 = 10kg, mass2 = 6kg) is measured when the objects are 12 centimeters apart. I

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Since the new distance is 3 times the old distance,
the new force is (1/3²) = 1/9th of the old force.

That's kind-of Choice-D, but I really don't like the way choice-D is worded.
"9 times smaller" is really pretty meaningless.

8 0

2 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:

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

$x_{1} = 0\\h_{s} = 4ft$

$F = \frac{(62.5)(100)(2)}{2}(2(0)+4)$

$F =25000lb$

2) Force on the triangular part:

$F = \frac{pg(l.h)}{6} (3x_{1} +2h)$

here

h = h(d) - h(s)

h = 10-4

h = 6ft

$x_{1} = 4ft\\$

$F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))$

$F = 150000 lb$

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

d) Force on the bottom:

$F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s}) }{2}$

$F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}$

$F = 2187937.5 lb$

7 0

3 years ago