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V125BC [204]
3 years ago
14

The temperature of the cosmic background radiation is measured to be 2.7 k. What is the wavelength of the peak in the spectral d

istribution at this temperature?
Physics
1 answer:
KATRIN_1 [288]3 years ago
7 0

Answer:

1.07\cdot 10^{-3} m

Explanation:

The peak wavelength of the spectral distribution can be found by using Wien's displacement law:

\lambda=\frac{b}{T}

where

b=2.898\cdot 10^{-3} m\cdot K is Wien's displacement constant

T is the absolute temperature

For the cosmic background radiation, the temperature is

T = 2.7 K

So, the corresponding peak wavelength is

\lambda=\frac{2.898\cdot 10^{-3} m\cdot K}{2.7 K}=1.07\cdot 10^{-3} m

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It would be 3.15 in meters
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Traveler A starts from rest at a constant acceleration of 6 m/s^2. Two seconds later, traveler B starts with an initial velocity
Troyanec [42]

Answer:

3. 3.5 s

Explanation:

The position of traveller A is given by the equation:

x_A(t) = \frac{1}{2}a t^2

where

a = 6 m/s^2 is the acceleration of A

t is the time measured from when A started the motion

The position of traveller B instead is given by

x_B(t) = u_B (t-2) + \frac{1}{2}a(t-2)^2

where a (acceleration) is the same as traveller A, and

u_B = 20 m/s

is B's initial velocity. We can verify that the formula is correct by substituting t=2, and we get x_B=0, which means that B starts its motion 2 seconds later.

Traveller B overtakes traveller A when the two positions are the same, so:

x_A = x_B\\\frac{1}{2}at^2 = u_B (t-2) + \frac{1}{2}a(t-2)^2\\\frac{1}{2}at^2 = u_B t - 2u_B +\frac{1}{2}at^2 +2a-2at\\u_Bt-2at = 2u_B-2a\\t=\frac{2u_B-2a}{u_B-2a}=\frac{2(20)-2(6)}{20-2(6)}=3.5 s

4 0
3 years ago
Most of the currents identified have a circular shape. This is because as the air and water move between the equator to the pole
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I believe it is true . Somebody correct me if I’m wrong!
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2 years ago
An object is thrown straight up into the air with an initial speed of 8 m/s, and reaches a greatest height of 15 m before it fal
Ugo [173]

Answer:

The value is T_t =  2.5659 \  s    

Explanation:

From the we are told that  

         The initial speed of the object is  u =  8 \  m/s

         The greatest height it reached is  h  =  15 \  m

Generally from kinematic equation we have that

      v^2 =  u^2 + 2gH

At maximum height v  =  0 m/s

So

      0^2 =  8^2 + 2 *  - 9.8 *  H

=>    H  =  3.27 \  m

Here H is the height from the initial height to the maximum height

So the initial height is mathematically represented as  

      s =  h - H

=>    s =  15 - 3.27

=>    s =  11.73 \  m

Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

            v  =  u + (-g) t

At maximum height v  =  0 m/s

           0 = 8 - 9.8t

=>         t = 0.8163 \  s

Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

       h  =  ut_1 + \frac{1}{2}  gt_1^2

Here the initial velocity is  0 m/s given that its the velocity at maximum height

Also  g is positive because we are moving in the direction of gravity  

So

       15  =  0* t  +  4.9 t^2

=>      t_1  =  1.7496

Generally the total time taken is mathematically represented as

          T_t =  t + t_1

=>        T_t =  0.8163  +   1.7496

=>        T_t =  2.5659 \  s            

 

6 0
3 years ago
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svet-max [94.6K]

a) Solids keep shape, liquids take shape of containers but don't spill, gases take container's shape and spill out

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c) Heat flows from warmer to colder bodies

d) For monatomic gases it's U=1.5nRT only, molecular gas has bonds between atoms so total internal energy increases

e) Of gases

f) v=\sqrt{\frac{3RT}{M}}=321 m/s

g) U=5/2*nRT=37830.85 J

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2 years ago
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