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Sunny_sXe [5.5K]
3 years ago
5

What are

Physics
1 answer:
koban [17]3 years ago
4 0
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What are the characteristics of a blackbody radiator? Give an example.
mihalych1998 [28]

Answer:

A blackbody, or Planckian radiator, is a cavity within a heated material from which heat cannot escape. No matter what the material, the walls of the cavity exhibit a characteristic spectral emission, which is a function of its temperature.

Example:

Emission from a blackbody is temperature dependent and at high temperature, a blackbody will emit a spectrum of photon energies that span the visible range, and therefore it will appear white. The Sun is an example of a high-temperature blackbody.

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3 years ago
Click the links to open the resources below. These resources will help you complete the assignment. Once you have created your f
Paul [167]

Answer:

Explanation:

lesgse in no

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3 years ago
Paying off your debts will most likely
g100num [7]
I think the answer is d
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3 years ago
39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
tatuchka [14]

<u>Answer:</u> The final temperature of the solution is 80.14^oC

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

T_{final}=80.14^oC

Hence, the final temperature of the solution is 80.14^oC

4 0
4 years ago
n a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic
Annette [7]

Answer:

Explanation:

a ) starting from rest , so u = o and initial kinetic energy = 0 .

Let mass of the skier = m

Kinetic energy gained = potential energy lost

= mgh = mg l sinθ

= m x 9.8 x 70 x sin 30

= 343 m

Total kinetic energy at the base = 343 m  + 0 = 343 m .

b )

In this case initial kinetic energy = 1/2 m v²

= .5 x m x 2.5²

= 3.125 m

Total kinetic energy at the base

= 3.125 m  + 343 m

= 346.125 m

c ) It is not surprising as energy gained due to gravitational force by the earth is enormous . So component of energy gained due to gravitational force far exceeds the initial kinetic energy . Still in a competitive event , the fractional initial kinetic energy may be the deciding factor .

7 0
4 years ago
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