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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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34kurt

Answer:

No. Because it would correspond to zero Instantaneous acceleration.

Explanation:

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What happens if the breakdown voltage is exceeded.
s2008m [1.1K]

Answer:

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Explanation:

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7 0
2 years ago
if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu
RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

F = \frac{k|q_1||q_2|}{r^2}

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N

Therefore, the mutual force between the two point charges is 319.64 N

4 0
3 years ago
A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.727 g, q = 2.73 µC is locat
Montano1993 [528]

Answer:

Q = 12.466μC

Explanation:

For the particle to execute a circular motion, the electrostatic force must be equal to the centripetal force:

Fe = \frac{K*q*Q}{r^{2}} = \frac{m*V^{2}}{r}

Solving for Q:

Q = \frac{m*V^{2}*r}{K*q}

Taking special care of all units, we can calculate the value of the charge:

Q = 12.466μC

5 0
3 years ago
To understand the terms in Faraday's law and to be able to identify the magnitude and direction of induced emf. Faraday's law st
wlad13 [49]

Answer: V_{\epsilon}\propto \frac{d\phi_{B}}{dt}

Explanation:

A direct proportionality means a linear relationship between two variables and rate of change means an application of derivatives. Hence, the mathematical model is:

V_{\epsilon}\propto \frac{d\phi_{B}}{dt}

5 0
3 years ago
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