1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
emmainna [20.7K]
2 years ago
10

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:
jenyasd209 [6]2 years ago
5 0
Sooooooooooo I’m pretty sure the answer is b
rodikova [14]2 years ago
4 0

Answer:

B I am not sure about the answer but it may be B

You might be interested in
A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through
Neporo4naja [7]

Answer:

Explanation:

Let the charge on proton be q .

energy gain by proton in a field having potential difference of V₀

= V₀ q

Due to gain of energy , its kinetic energy becomes 1/2 m v₀²

where m is mass and v₀ is velocity of proton

V₀ q = 1/2 m v₀²

In the second case , gain of energy in electrical field

= 2 V₀q , if v be the velocity gained in the second case

2 V₀q = 1/2 m v²

1/2 m v² = 2 V₀q = 2 x 1/2 m v₀²

mv² = 2  m v₀²

v = √2 v₀

6 0
2 years ago
HELP HELP HELP help pls
GalinKa [24]
Answer is A.) 40 m/km

7 0
2 years ago
A “point-light walker” wears lights on different body locations. When viewed in a dark room, an observer would perceive a(n):
lions [1.4K]

Answer:

person when the point-light walker is moving

Explanation:

A point light walker is an arrangement of dots that moves in a way that mimics a human walking. This is used in the field of Biological Motion Perception.

Biological motion perception is the science that deals with how our brain perceives motion. In order to understand how the brain perceives motion a point light walker is used.

6 0
2 years ago
Sound is faster than light that is why we hear thunder before we see lightning
Umnica [9.8K]

Answer:

false

Explanation:

sound travels slower than light. that is why we see lightning before we hear the thunder

8 0
3 years ago
A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput
Nikitich [7]
There is a missing data in the text of the problem (found on internet):
"with force constant<span> k=</span>450N/<span>m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
</span>E=U+K=  \frac{1}{2}kx^2 + \frac{1}{2} mv^2
<span>where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
</span>E=K_{max} =  \frac{1}{2}mv_{max}^2
<span>Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
</span>E=U_{max}= \frac{1}{2}k A^2
<span>
Since the mechanical energy must be conserved, we can write
</span>\frac{1}{2}mv_{max}^2 =  \frac{1}{2}kA^2
<span>from which we find the maximum speed
</span>v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }=  1.2 m/s
<span>
b) </span><span> the </span>speed<span> of the </span>glider<span> when it is at x= -0.015</span><span>m

We can still use the conservation of energy to solve this part. 
The total mechanical energy is:
</span>E=K_{max}=  \frac{1}{2}mv_{max}^2= 0.36 J
<span>
At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
</span>U= \frac{1}{2}kx^2 =  \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J
<span>And since 
</span>E=U+K
<span>we find the kinetic energy when the glider is at this position:
</span>K=E-U=0.36 J - 0.05 J = 0.31 J
<span>And then we can find the corresponding velocity:
</span>K= \frac{1}{2}mv^2
v=  \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s
<span>
c) </span><span>the magnitude of the maximum acceleration of the glider;
</span>
For a simple harmonic motion, the magnitude of the maximum acceleration is given by
a_{max} = \omega^2 A
where \omega= \sqrt{ \frac{k}{m} } is the angular frequency, and A is the amplitude.
The angular frequency is:
\omega =  \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s
and so the maximum acceleration is
a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2

d) <span>the </span>acceleration<span> of the </span>glider<span> at x= -0.015</span><span>m

For a simple harmonic motion, the acceleration is given by
</span>a(t)=\omega^2 x(t)
<span>where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find 
</span>a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2
<span>
e) </span><span>the total mechanical energy of the glider at any point in its motion. </span><span>

we have already calculated it at point b), and it is given by
</span>E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J
8 0
3 years ago
Other questions:
  • What is the average kinetic energy in joules of hydrogen atoms on the 5500ºc surface of the sun?
    7·1 answer
  • How does the theory of natural selection explain the diversity of life?
    10·1 answer
  • A physicist is calibrating a spectrometer that uses a diffraction grating to separate light in order of increasing wavelength (λ
    6·1 answer
  • When current flows through an electric motor, the electromagnet rotates. true or false?
    5·1 answer
  • A 85.4 kg ice skater is moving at 5.97 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole.
    12·1 answer
  • A 30.0 kg object rests on a flat, frictionless surface. A rope lifts up on the object with a force of 309 N. What is the acceler
    7·1 answer
  • You pull a 200 kg box straight up with a rope connected to a helicopter. If the
    5·2 answers
  • A gazelle is running at 17.46 m/s. He sees a lion and accelerates at -1.49 m/s/s,
    15·2 answers
  • Whats a good string length for a parachute
    6·1 answer
  • Ramu,the gardener,is trying to pull out weeds. however,he has to apply great force.why do you think he has to apply to much forc
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!