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PtichkaEL [24]
3 years ago
14

Consider a blackbody that radiates with an intensity i1 at a room temperature of 300k. At what intensity i2 will this blackbody

radiate when it is at a temperature of 400k?
Physics
1 answer:
alexgriva [62]3 years ago
4 0

Answer:

3.16 i1

Explanation:

T1 = 300 k

I1 = i1

T2 = 400 k

i2 = ?

The intensity is directly proportional to the four powers of the absolute temperature.

I ∝ T^4

So, \frac{i_{2}}{i_{1}}=\frac{T_{2}^{4}}{T_{1}^{4}}

\frac{i_{2}}{i_{1}}=\frac{400^{4}}{300^{4}}

i2 = 3.16 i1

Thus, the intensity becomes 3.16 i1.

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You must observe the object twice.

-- Look at it the first time, and make a mark where it is.

-- After some time has passed, look at the object again, and
make another mark at the place where it is.

-- At your convenience, take out your ruler, and measure the
distance between the two marks.

What you'll have is the object's "displacement" during that period
of time ... the distance between the start-point and end-point. 
Technically, you won't know the actual distance it has traveled
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8 0
3 years ago
When you have a pot of water on the stove, heat is transferred to the water. Describe the behavior of the water molecules and ho
Digiron [165]

The hotter molecules become, the faster they move around. The colder they are, the more slow and lethargic they are

3 0
3 years ago
Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 50 W/m · K and
tatuchka [14]

Answer:

solution:

dT/dx =T2-T1/L

&

q_x = -k*(dT/dx)

<u>Case (1)  </u>

dT/dx= (-20-50)/0.35==> -280 K/m

 q_x  =-50*(-280)*10^3==>14 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

 q_x  =-50*(80)*10^3==>-4 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

 q_x  =-50*(80)*10^3==>-4 kW

Case (3)

q_x  =-50*(160)*10^3==>-8 kW

T2=T1+dT/dx*L=70+160*0.25==> 110° C

Case (4)

q_x  =-50*(-80)*10^3==>4 kW

T1=T2-dT/dx*L=40+80*0.25==> 60° C

Case (5)

q_x  =-50*(200)*10^3==>-10 kW

T1=T2-dT/dx*L=30-200*0.25==> -20° C

note:

all graph are attached

6 0
3 years ago
What is the force needed to move an 225 kg car a distance of 150 meters
Kobotan [32]

Physics

mass = 225 kg

weight = m × a = 2250 N

distance = 150 m

force : ______?

Answers :

W = F × s

W = 2250 × 150

W = 337500 ✅

4 0
3 years ago
Read 2 more answers
I need help with this answer
7nadin3 [17]

Answer:

double replacement

Explanation:

sorry if im wrong

8 0
3 years ago
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