All categories

2 years ago

Real bodies emit and absorb more radiation than a blackbody at the same temperature. True or False

Physics

1 answer:

monitta2 years ago

3 0

Answer: False

Explanation:

Relative to the concept of radiations, a black body is an object capable of absorbing any form of electromagnetic radiation irrespective of its frequency or angle of incidence when incident on such object.

However, the same cannot be said about real bodies as real bodies are those which reflect all rays incident on them completely and uniformly in all directions.

One very important characteristic of black bodies is that they are ideal emmiters.

The concept of emmisivity is brought about by the existence of real bodies .

This is due to the fact that they are only able to emit radiation at a fraction of the black body energy levels.

Please note that by convention, the emmisivity of a real body is always less thaan 1.

As such they are not able to emit as much radiation as a black body at the same temperature.

You might be interested in

Does voltage remain the same throughout a series circuit

seropon [69]

It depends on the circuit. Sometimes it becomes a bit weaker, sometimes it stays the same.

6 0

2 years ago

Read 2 more answers

Drawing a ray diagram for an object far from convex lens requires how many rays?

Crank

Answer

Explanation:

four rays can also explain the ray diagram if the object is at infinity

5 0

2 years ago

The normal force acting on an object and the force of static friction do zero work on the object. However the reason that the wo

spin [16.1K]

Answer:

<em>The normal force is perpendicular to the displacement</em>

<em>The static friction force produces no displacement</em>

Explanation:

Work Done By Special Forces

The work is a physical magnitude that measures the dot product of the force applied to an object by the displacement it produces in it.

$W=\vec F\ \vec r$

It can be written in its scalar version as

$W=F.d.cos\theta$

Being F and d the magnitudes of the force and displacement, and $\theta$ the angle between them

If the angle is zero, the work is at maximum, it the angle is 90°, the work is zero. If the angle is between 90° and 180°, the work is negative.

The normal force acts in the vertical direction when the object is being pushed horizontally. It means the angle between the force and the displacement is 90°, thus the work is

$W=N.d.cos90^o=0$

The work is zero because the force and the displacement are perpendicular

The static friction force exists only when the object is being applied a force of a magnitude not large enough to produce movement, i.e. the object is at rest. If the object is moved, the friction force is still present, but it's called dynamic friction force, usually smaller than the static.

Since in this case, there is no displacement, d=0, and the work is

$W=F_r(0)cos180^o=0$

3 0

3 years ago

Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$

The speed of A is $v__{A}} = 20 \ m/s$

The uniform acceleration of B is $a__{B}} = 0.333 \ m/s^2$

Generally the time taken by go-cart A is mathematically represented as

$t__{A}} = \frac{l}{v__{A}}}$

=> $t__{A}} = \frac{1000}{20}$

=> $t__{A}} = 50 \ s$

Generally from kinematic equation we can evaluate the time taken by go-cart B as

$l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest u = 0 m/s

$1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=> $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$

=> $t__{B}} = 77.5 \ seconds$

Comparing $t__{A}} \ and \ t__{B}}$ we see that $t__{A}}$ is smaller so go-cart A is faster

3 0

2 years ago

Dans un tube en U contenant du mercure ,on verse de l'autre côté de l'acide sulfurique de densité 1,84 et de l'autre côté de l'a

Iteru [2.4K]

Explanation:

Unclear question. The clear rendering reads;

"Into a U-tube containing mercury, pour on the other side sulfuric acid of density 1.84 and on the other side alcohol of density 0.8 so that the levels are in the same horizontal plane. The height of the acid above the mercury being 24 cm. What is the height of the bar and what variation of the level of the acid, when the mercury density is 13.6?

6 0

2 years ago