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MA_775_DIABLO [31]
3 years ago
15

What is responsible for the production of x ray emission at the cutoff wavelength?

Physics
1 answer:
Katen [24]3 years ago
3 0

Answer:

correct answer is option 5 (The emission by the incident electron when it loses its energy in a single emission)

Explanation:

The emission by the incident electron when it loses its energy in a single emission then X-rays of minimum wavelength which is also known as X-rays of cut-off wavelength are emitted.

If potential V is applied on coolidge tube then cut-off wavelength of the continuous X-ray spectrum is related to the accelerating potential  V   by following equation

\lambda_{cut-off} =\frac{hc}{eV} =\frac{12400}{V} Angstrom/V

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In the United States, household electric power is provided at a frequency of 60 HzHz, so electromagnetic radiation at that frequ
grigory [225]

Answer:

the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

Explanation:

Given the data in the question;

To determine the maximum intensity of an electromagnetic wave, we use the formula;

I = \frac{1}{2}ε₀cE_{max²

where ε₀ is permittivity of free space ( 8.85 × 10⁻¹² C²/N.m² )

c is the speed of light ( 3 × 10⁸ m/s )

E_{max is the maximum magnitude of the electric field

first we calculate the maximum magnitude of the electric field ( E_{max  )

E_{max = 350/f kV/m

given that frequency of 60 Hz, we substitute

E_{max = 350/60 kV/m

E_{max = 5.83333 kV/m

E_{max = 5.83333 kV/m × ( \frac{1000 V/m}{1 kV/m} )

E_{max = 5833.33 N/C

so we substitute all our values into the formula for  intensity of an electromagnetic wave;

I = \frac{1}{2}ε₀cE_{max²

I = \frac{1}{2} × ( 8.85 × 10⁻¹² C²/N.m² ) × ( 3 × 10⁸ m/s ) × ( 5833.33 N/C )²

I = 45 × 10³ W/m²

I = 45 × 10³ W/m² × ( \frac{1 kW/m^2}{10^3W/m^2} )

I = 45 kW/m²

Therefore, the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

7 0
3 years ago
Two parallel conducting plates are connected to a constant voltage source. The magnitude of the electric field between the plate
Alecsey [184]

Answer:

The magnitude of the new electric field is <u>35820 N/C</u>.

Explanation:

Given:

Original magnitude of electric field (E₀) = 2388 N/C

Original voltage = 'V' (Assume)

Original separation between the plates = 'd' (Assume)

Now, new voltage is three times original voltage. So, V_n=3V

New distance is 1/5 the original distance. So, d_n=\dfrac{d}{5}

Now, electric field between the parallel plates originally is given as:

E_0=\frac{V}{d}=2388\ N/C

Let us find the new electric field based on the above formula.

E_n=\frac{V_n}{d_n}\\\\E_n=\frac{3V}{\frac{d}{5}}\\\\E_n=15(\frac{V}{d})

Now, \frac{V}{d}=2388\ N/C. So,

E_n=15\times 2388=35820\ N/C

Therefore, the magnitude of the new electric field is 35820 N/C.

3 0
3 years ago
A man wears convex lens glasses of focal length 30cm in order to correct his eyes defect. Instead of the optimum 25cm, his dista
omeli [17]

Answer:

14 cm

Explanation:

F = (frac{uv}{u – v})

F = +ve

v = -ve

30 = (frac {25 {times} (-v)}{25 – (-v)})

v = (frac {25 {times} (-v)}{25+v})

v = 14cm

(Note that either negative or positive values go to show the positioning and hence, they are not a strong necessity in your final answer.)

So happy that i could help you!

Now this question could turn out to be easy for you!!

7 0
2 years ago
When performing an.experiment similar to Millikan's oil drop, a student measured the following load magnitudes: 3.26x10 ^-19 C 5
torisob [31]

Answer:

Explanation:

We put the charges in the ascending order as follows

1.53 P

3.26 P

4.66 P

5.09 P

6.39 P

where P is equal to 10⁻¹⁹

we round off given charges as follows

1.53 P → 1.6 P

3.26 P → 3.2 P

4.66 P → 4.8 P

5.09 P → 4.8 P

6.39 P → 6.4 P

We see that 2 nd to 4 th charges are integral multiples of first charge . That means these charges are supposed to be made of combination of first charge . So first charge appears to be minimum possible charge .

Hence this charge may exist on single electron.

8 0
3 years ago
Which of the circuit offers greater resistance to the flow of current 1A or 2A ?
Dafna1 [17]
1 A offers greater resistance
7 0
3 years ago
Read 2 more answers
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