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saveliy_v [14]
3 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]3 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]3 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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Given the height of rod,length of shadow of tree and length of shadow of the rod, estimate the height of the tree? Given:height
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Answer:

1000 cm.

Explanation:

To obtain the estimated tree height :

(Height of rod / length of rod shadow) = (height of tree / length of tree shadow)

Substituting values into the formula :

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Using cross multiplication :

Height of tree * 120 = 150 * 800

Height of tree = (150 * 800) / 120

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Describe the importance of conservative forces to conservation of energy.
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If the mass of an object is 35kg on venus what is its mass on the sun
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8 0
3 years ago
A luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 25 degrees above the horizontal by a force F of magnitude 145 N
Nonamiya [84]

Answer:

A) 667 J

B) 381.4 J

C) 0 J

D) 245.4 J

E) 40.2J

F) 2 m/s

Explanation:

Let g = 9.81 m/s2

A) The work done on the suitcase is the product of the force applied and the distance travelled:

w = Fs = 145 * 4.6 = 667 J

B) The work done by gravitational force the dot product between the gravity vector and the distance vector

W_g = \vec{P}\vec{s} = mgs sin\alpha = 20*9.81*4.6*sin25^o = 381.4 J

C) As the normal force vector is perpendicular to the distance vector, the work done by the normal force is 0

D) The work done on the suitcase by friction force is the product of the force applied and the distance travelled, whereas friction force is the product of normal force and coefficient

W_f = F_fs = \mu N s = \mu s mgcos\alpha = 0.3* 4.6 * 20*9.81*cos25^o = 245.4 J

E) The total workdone on the suite case would be the pulling work subtracted by gravity work and friction work

W = w – W_g – W_f = 667 – 381.4 – 245.4 = 40.2 J

F) As the suit case has 0 kinetic and potential energy at the bottom, and the total work done is converted to kinetic energy at 4.6 m along the ramp, we can conclude that:

E_k = W = 40.2 j

mv^2/2 = 40.2

20v^2/2 = 40.2

10v^2 = 40.2

v^2 = 4.02

v = \sqrt{4.02} = 2 m/s  

3 0
3 years ago
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