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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 earth exerts a force of 1.00 newtons on an object in free fall. what is the objects mass?
ICE Princess25 [194]
On or near the surface of the Earth, 1 newton is the weight of about 102 grams of mass.

Note that the gravitational force between the object and the Earth is always the
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3 0
3 years ago
A cylindrical barrel is completely full of water and sealed at the top except for a narrow tube extending vertically through the
astraxan [27]

To solve this problem it is necessary to apply the concepts related to pressure as a unit that measures the force applied in a specific area as well as pressure as a measurement of the density of the liquid to which it is subjected, its depth and the respective gravity.

The two definitions of pressure can be enclosed under the following equations

P = \frac{F}{A}

Where

F= Force

A = Area

P = \rho gh

Where,

\rho = Density

g = Gravity

h = Height

Our values are given as,

d = 0.8m \rightarrow r = 0.4m

A = \pi r^2 = \pi * 0.4^2 = 0.503m^2

If we make a comparison between the lid and the tube, the diameter of the tube becomes negligible.

Matching the two previous expressions we have to

\frac{F}{A} = \rho g h

Re-arrange to find h

h = \frac{F}{A\rho g}

h = \frac{390}{(0.503)(1000)(9.8)}

h = 0.079m

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8 0
3 years ago
iven a 36.0 V battery and 14.0 Ω and 84.0 Ω resistors, find the current (in A) and power (in W) for each when connected in serie
levacccp [35]

Answer:

0.367A = Current of both resistors

For resistor 1: 1.89W; For resistor 2: 11.3W

Explanation:

When the resistors are connected in series, the equivalent resistance is the sum of both resistors, that is:

R = 14.0Ω + 84.0Ω = 98.0Ω

Using Ohm's law, we can find the current of the circuit (Is the same for both resistors):

V = RI

V / R = I

36.0V / 98.0Ω = I

<h3>0.367A = Current of both resistors</h3><h3 />

Power is defined as:

P = I²*R

For resistor 1:

P = 0.367A²*14.0Ω = 1.89W

For resistor 1:

P = 0.367A²*84.0Ω = 11.3W

6 0
3 years ago
What is the name for a star and planets held together by gravity? A. solar system B. galaxy C. black hole D. supernova
DaniilM [7]
These answers aren´t valid .

The correct answer will be:

Planetary system.
5 0
4 years ago
A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c
netineya [11]

Answer:

E=\frac{\lambda}{2\pi r\epsilon_0}

Explanation:

We are given that

Linear charge density of wire=\lambda

Radius of hollow cylinder=R

Net linear charge density of cylinder=2\lambda

We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R

By Gauss theorem

\oint E.dS=\frac{q}{\epsilon_0}

q=\lambda L

E(2\pi rL)=\frac{L\lambda}{\epsilon_0}

Where surface area of cylinder=2\pi rL

E=\frac{\lambda}{2\pi r\epsilon_0}

8 0
3 years ago
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