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larisa86 [58]
4 years ago
6

The Cosmic Background Radiation Outer space is filled with a sea of photons, created in the early moments of the universe. The f

requency distribution of this "cosmic background radiation" matches that of a blackbody at a temperature near 2.7K.What is the peak frequency of this radiation in Hz?What is the wavelength that corresponds to the peak frequency in mm?
Physics
1 answer:
lions [1.4K]4 years ago
8 0

A) Peak wavelength: 1.07 mm

The peak wavelength of the Cosmic Background Radiation can be found by using Wien's displacement law:

\lambda = \frac{b}{T}

where

\lambda is the peak wavelength

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

T is the absolute temperature

For the Cosmic Background Radiation,

T = 2.7 K

So the peak wavelength is

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

B) Peak frequency: 2.8\cdot 10^{11}Hz

The peak frequency can be found by using the relationship:

f=\frac{c}{\lambda}

where

c=3.0\cdot 10^8 m/s is the speed of light

\lambda is the peak wavelength

Substituting numbers, we find

f=\frac{3.0\cdot 10^8 m/s}{1.07\cdot 10^{-3} m}=2.8\cdot 10^{11}Hz

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Answer:

(a) v_{f}=26.08m/s

(b) t=4.906seconds

Explanation:

We will take ground level as origin and upward is positive direction

the givens are

y_{i}=0\\y_{f}=-9.0m\\ v_{i}=22m/s

Part (a)

To find the final velocity we use the kinematic equation

So

(v_{f})^{2} =(v_{i})^{2} +2a(y_{f}-y_{i})\\(v_{f})^{2}=(22m/s)^{2}+2(-9.8m/s^{2} ) (-10-0)\\v_{f}=\sqrt{(22m/s)^{2}+2(-9.8m/s^{2} ) (-10-0)} \\v_{f}=26.08m/s

Part (b)

To find time of rock trip until it touches the ground we will use simple kinematic equation or simple motion equation

v_{f}=v_{i}+at\\t=\frac{v_{f}-v_{i}}{a}\\ t=\frac{(-26.08m/s)-(22m/s)}{-9.8m/s^{2} }\\t=4.906seconds

Notice that we substituted vf with negative sign because its direction is downwards

5 0
3 years ago
3. If a pipe with flowing water has a cross-sectional area nine times greater at point 2 than at point 1, what would be the rela
shusha [124]

Answer:

The velocity of the flow at point 1 is nine times greater than the velocity at point 2.

Explanation:

We know that the same volume of water entering through point 1 must exit through point 2.

The volume of water per unit of time or the volumetric flow is defined by:

V_{flow}=v*A\\ where:\\V_{flow}= flow [m^3/s]\\v=velocity[m/s]\\A=area [m^2]

If  A_{2}=9*A_{1}  \\

therefore

v_{1} *A_{1} = v_{2} *A_{2} \\v_{1} *A_{1} = v_{2} *9A_{1} \\\\v_{1} = 9v_{2}  \\

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In a container of negligible mass, 0.400 kg of ice at an initial temperature of -29.0 ∘C is mixed with a mass m of water that ha
Pie

Answer:

1 kg

Explanation:

The container has negligible mass and no heat is loss to the surrounding.

Mass of ice = 0.4kg, initial temperature of ice = -29oC, final temperature of the mixture = 26oC, mass of water (m2) = ?kg, initial temperature of water = 80oC, c ( specific heat capacity of water ) = 4200J/kg.K, Lf = heat of fusion of water = 3.36 × 10^5 J/kg

Using the formula:

Quantity of heat gain by ice = Quantity of heat loss by water

Quantity of heat gain by ice = mass of ice × heat of fusion of ice + mass of water × specific heat capacity of water = (0.4 × 3.36 × 10^ 5) + (0.4 × 4200 × (26- (-29) = 13.44 × 10^4 + 9.24 × 10^ 4 = 22.68 × 10^4 J

Quantity of heat loss by water = m2cΔT

Quantity of heat loss by water = m2 ×4200× (80 - 26) = m(226800)

since heat gain = heat loss

22.68 × 10^4 = 226800 m2

divide both side by 226800

226800 / 226800 = m2

m2 = 1 kg

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Answer:

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c. Orbit of geostationary satellite

d. Orbit of the earth.

Explanation:

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