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saveliy_v [14]
2 years ago
8

When an object radiates heat, the strength of this radiation far from the object decreases when distance from the source increas

es as shown in the graph below:
That is, radiated heat is much stronger near its source.

The universe is full of heat that was radiated by a source that no longer exists. This heat is known as cosmic background radiation. Cosmic background radiation is not stronger in any one direction or part of the universe than in others.

The following image is a map of the cosmic background radiation. Red areas are only 0.0002 K hotter than the blue areas. The overall temperature of the radiation is 2.725 K.


Image by the WMAP team, courtesy of the Legacy Archive
for Microwave Background Data Analysis (LAMBDA) supported by NASA

What does the uniformity of this radiation imply about its source?
A.
The source of cosmic background radiation existed for a very long time.
B.
The source of cosmic background radiation existed for a very short time.
C.
The source of cosmic background radiation moved randomly.
D.
The source of cosmic background radiation filled the entire universe.
Physics
2 answers:
Andrew [12]2 years ago
7 0

Answer:

The source of cosmic background radiation filled the entire universe.

Explanation:

D:The source of cosmic background radiation filled the entire universe.

miss Akunina [59]2 years ago
3 0

Answer:

Ur answer is D:The source of cosmic background radiation filled the entire universe.

Explanation:

Hope this helps.

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Iron and stone

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Where does the rest of the energy 90% go?
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An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu
Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

\alpha = \frac{\omega_f - \omega_i}{t}

where we have

\alpha = -80.0 rad/s^2 is the angular acceleration (negative since the rotor is slowing down)

\omega_f is the final angular speed

\omega_i = 2000 rad/s is the initial angular speed

t = 10.0 s is the time interval

Solving for \omega_f, we find the final angular speed after 10.0 s:

\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

\alpha = \frac{\omega_f - \omega_i}{t}

If we re-arrange it for t, we get:

t = \frac{\omega_f - \omega_i}{\alpha}

where here we have

\omega_i = 2000 rad/s is the initial angular speed

\omega_f=0 is the final angular speed

\alpha = -80.0 rad/s^2 is the angular acceleration

Solving the equation,

t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s

6 0
3 years ago
A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853
Norma-Jean [14]

Recall that

{v_f}^2-{v_i}^2=2a\Delta x

where v_i and v_f are the initial and final velocities, respecitvely; a is the acceleration; and \Delta x is the change in position.

So we have

\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)

\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}

(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)

7 0
3 years ago
A test tube of length L and cross-sectional area A is submerged in water with the open end down so that the edge of the tube is
Margaret [11]

Answer:

if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

Explanation:

The air in the tube can be considered an ideal gas,

           P V = nR T

In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H

For pressure the open end of the tube is

         P₂ = P_atm + ρ g H

Let's write the gas equation for the colon

            P₁ V₁ = P₂ V₂

            P_atm V₁ = (P_atm + ρ g H) V₂

             V₂ = V₁    P_atm / (P_atm + ρ g h)

If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

The main assumption is that the temperature during the experiment does not change

6 0
3 years ago
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