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mafiozo [28]
4 years ago
7

What idea did Max Planck propose to help explain why a blackbody radiator did not give off light of increasingly high frequency

as its temperature increased?
A. Matter behaves as particles under certain conditions.

B. Matter can absorb energy in any amount.

C. Matter can absorb light only in certain specific amounts.

D. Matter behaves as a wave under certain conditions.
Physics
2 answers:
Leno4ka [110]4 years ago
5 0
The idea that <span>Max Planck propose to help explain why a blackbody radiator did not give off light of increasingly high frequency as its temperature increased is that </span>C. Matter can absorb light only in certain specific amounts. 
Fiesta28 [93]4 years ago
4 0

Answer:

C). Matter can absorb light only in certain specific amounts.

Explanation:

We know that any objects that is heated give out radiation in the form of energy. Classical physics cannot be used to describes black body radiation. Thus Max Plank suggested that had to absorb energy in small packets of energy or in discrete steps. That is matter can not absorb any amount of energy, but it only does in certain amount of energy called the photons.

The energy absorbed or released can be expressed as

ΔE = n.h.v

where n is whole number 1, 2, 3, 4, 5...

h is Planks constant ( 6.626 x 10^{-34} J-s

v is light energy absorbed or released

Hence, matter can absorb light energy only in certain specific amounts.

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The velocity profile in fully developed laminar flow in a circular pipe of inner radius R 5 2 cm, in m/s, is given by u(r) 5 4(1
xxMikexx [17]

The question is not clear and the complete clear question is;

The velocity profile in fully developed laminar flow In a circular pipe of inner radius R = 2 cm, in m/s, is given By u(r) = 4(1 - r²/R²). Determine the average and maximum Velocities in the pipe and the volume flow rate.

Answer:

A) V_max = 4 m/s

B) V_avg = 2 m/s

C) Flow rate = 0.00251 m³/s

Explanation:

A) We are given that;

u(r) = 4(1 - (r²/R²))

To obtain the maximum velocity, let's apply the maximum condition for a single-variable continual real valued problem to obtain;

(d/dr)(u(r)) = 0

Thus,

(d/dr)•4(1 - (r²/R²)) = 0

4(d/dr)(1 - (r²/R²)) = 0

If we differentiate, we have;

4(0 - (2r/R²)) = 0

-8r/R² = 0

Thus, r = 0 and with that, the maximum velocity is at the centre of the pipe.

Thus, for maximum velocity, let's put 0 for r in the U(r) function.

Thus,

V_max = 4(1 - 0²/R²) = 4 - 0 = 4 m/s

B) Average velocity is given by;

V_avg = V_max/2

V_avg = 4/2 = 2 m/s

C) the flow can be calculated from;

Flow rate ΔV = A•V_avg

A is area = πr²

From question, r = 2cm = 0.02m

A = π x 0.02²

Hence,

ΔV = π x 0.02² x 2 = 0.00251 m³/s

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