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Kamila [148]
3 years ago
6

Two sources emit beams of light of wavelength 550 nm. The light from source A has an intensity of 10 μW/m2, and the light from s

ource B has an intensity of 20 μW/m2. This is all we know about the two beams. Which of the following statements about these beams are correct? A) Beam B carries twice as many photons per second as beam A. B) A photon in beam B has twice the energy of a photon in beam A. C) The frequency of the light in beam B is twice as great as the frequency of the light in beam A. D) A photon in beam B has the same energy as a photon in beam A. E) None of the above statements are true.
Physics
1 answer:
alexira [117]3 years ago
3 0

Answer:

<em>A) Beam B carries twice as many photons per second as beam A.</em>

Explanation:

If we have two waves with the same wavelength, then their intensity is proportional to their power, or the energy per unit time.

We also know that the amount of photon present in an electromagnetic beam is proportional to the energy of the beam, hence the amount of beam per second is proportional to the power.

With these two facts, we can say that the intensity is a measure of the amount of photon per second in an electromagnetic beam. So we can say that <em>beam B carries twice as more power than beam A, or Beam B carries twice as many photons per second as beam A.</em>

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Answer:

Magnetic field, B=2\times 10^{-4}\ T

Explanation:

Given that,

Velocity of electron, v=5\times 10^7\ m/s

It enters  a region of space where perpendicular electric and a magnetic fields are present.

Magnitude of electric field, E=10^4\ V/m

We need to find the magnetic field will allow the electron to go through the region without being deflected.

Magnetic force on the electron, F_m=qvB\ sin\theta.......(1)

Electric force on the electron, F = q E........(2)

From equation (1) and (2) we get:

qvB\ sin\theta=qE

B=\dfrac{E}{v}

B=\dfrac{10^4\ V/m}{5\times 10^7\ m/s}

B = 0.0002 T

or

B=2\times 10^{-4}\ T

Hence, this is the required solution.

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You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi
Mama L [17]

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

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3 years ago
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serg [7]
Its magnetic..........ig

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4 years ago
Steam enters a counterflow heat exchanger operating at steady state at 0.07 MPa with a specific enthalpy of 2431.6 kJ/kg and exi
PIT_PIT [208]

Answer:

Explanation:

Given:

Steam Mass rate, ms = 1.5 kg/min

= 1.5 kg/min × 1 min/60 sec

= 0.025 kg/s

Air Mass rate, ma = 100 kg/min

= 100 kg/min × 1 min/60 sec

= 1.67 kg/s

A.

Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.

xf, quality = 0.9.

Tsat = 89.9°C

hf = 376.57 kJ/kg

hfg = 2283.38 kJ/kg

Using the equation for specific enthalpy,

hi = hf + (hfg × xf)

= 376.57 + (2283.38 × 0.9)

= 2431.552 kJ/kg

The specific enthalpy of the outlet, h2 = hf

= 376.57 kJ/kg

B.

Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy

= ms × (hi - h2)

= 0.025 × (2431.552 - 376.57)

= 0.025 × 2055.042

= 51.37455 kW

= 51.38 kW.

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3 years ago
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