When a piece of metal is heated to high temperatures, it begins to glow red,

Physics

1 answer:

salantis [7]2 years ago

7 0

Answer:

The particle model, because the metal is only absorbing radiation

of specific frequencies

Explanation:

The heating of a metal is an example of blackbody radiation. Blackbody radiation refers to the spectrum of light emitted by any heated object; common examples include the heating element of a toaster and the filament of a light bulb. The spectral intensity of blackbody radiation peaks at a frequency that increases with the temperature of the emitting body(Encyclopedia Britianica).

Max Plank proffered explanation to black body radiation by assuming that the oscillators in the heated body emit or absorb energy in discrete frequencies given by E=hf. Hence the energy was directly proportional to the frequency of the oscillator.

You might be interested in

Copper has a specific heat of 0.09 cal/g x C . How much change in temperature would the addition of 8373 cal of heat have on a 5

neonofarm [45]

<h3>Answer:</h3>

172.92 °C

<h3>Explanation:</h3>

Concept being tested: Quantity of heat

We are given;

Specific heat capacity of copper as 0.09 cal/g°C
Quantity of heat is 8373 calories
Mass of copper sample as 538.0 g

We are required to calculate the change in temperature.

In this case we need to know that the amount of heat absorbed or gained by a substance is given by the product of mass, specific heat capacity and change in temperature.

That is, Q = m × c × ΔT

Therefore, to calculate the change in temperature, ΔT we rearrange the formula;

ΔT = Q ÷ mc

Thus;

ΔT = 8373 cal ÷ (538 g × 0.09 cal/g°C)

= 172.92 °C

Therefore, the change in temperature will be 172.92 °C

3 0

2 years ago

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated

Lelu [443]

Incomplete question as we have not told to find what quantity.The complete question is here

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 5.00 cm.calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

Answer:

(a) $C=16.7pF$

(b) $r_{a} =3.749cm$

Explanation:

Given data

$Q=3.50nC\\V=210V\\r_{b}=5.0cm$

For part (a)

The Capacitance given by:

$C=\frac{Q}{V}\\ C=\frac{3.50*10^{-9} C}{210V}\\C=1.6666*10^{-11}F\\or\\C=16.7pF$

For part (b)

The Capacitance of coordinates is given as

$C=\frac{4\pi e}{\frac{1}{r_{a} }-\frac{1}{r_{b} } }\\ So\\{\frac{1}{r_{a} }-\frac{1}{r_{b} } }=\frac{4\pi *8.85*10^{-12} }{1.666*10^{-11}}=6.672m^{-1} \\ \frac{1}{r_{a} }=6.672+(1 /0.05)\\\frac{1}{r_{a} }=26.672\\r_{a} =1/26.672\\r_{a} =0.0375m\\r_{a} =3.749cm$