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3 years ago

To understand the experiment that led to the discovery of the photoelectric effect.

Physics

1 answer:

andrew11 [14]3 years ago

6 0

Answer:

A) Emin = eV

B) Vo = (E_light - Φ) ÷ e

Explanation:

Energy of electron is the product of electron charge and the applied potential difference.

The energy of an electron in this electric field with potential difference V will be eV. Since this is the least energy that the electron must reach to break out, then the minimum energy required by this electron will be;

Emin = eV

The maximum stopping potential energy is eVo,

The energy of the electron due to the light is E_light.

If the minimum energy electron must posses is Φ, then the minimum energy electron must have to reach the detectors will be equal to the energy of the light minus the maximum stopping potential energy

Φ = E_light - eVo

Therefore,

eVo = E_light - Φ

Vo = (E_light - Φ) ÷ e

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A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy (k = 160 W/m-K). The fin diameter is D = 4 mm, and

frozen [14]

Answer:

Given that

D= 4 mm

K = 160 W/m-K

h=h = 220 W/m²-K

ηf = 0.65

We know that

$m=\sqrt{\dfrac{hP}{KA}}$

For circular fin

$m=\sqrt{\dfrac{4h}{KD}}$

$m=\sqrt{\dfrac{4\times 220}{160\times 0.004}}$

m = 37.08

$\eta_f=\dfrac{tanhmL}{mL}$

$0.65=\dfrac{tanh37.08L}{37.08L}$

By solving above equation we get

L= 36.18 mm

The effectiveness for circular fin given as

$\varepsilon =\dfrac{2\ tanhmL}{\sqrt{\dfrac{hD}{K}}}$

$\varepsilon =\dfrac{2\ tanh(37.08\times 0.03618)}{\sqrt{\dfrac{220\times 0.004}{160}}}$

ε = 23.52

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2 years ago

Read 2 more answers

A scaffold of mass 57 kg and length 6.5 m is supported in a horizontal position by a vertical cable at each end. A window washer

Mrac [35]

Explanation:

The given data is as follows.

$m_{1}$ = 57 kg, $m_{2}$ = 79 kg

$l_{1}$ = 6.5 m, $l_{2}$ = (6.5 - 1.9) m = 4.6 m

(a) The sum of torque ends about far end is as follows.

$m_{1}g \frac{l_{1}}{2} + m_{2}g \times l_{2} - T \times l_{1}$ = 0

$57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (6.5 - 1.9) - T \times 6.5$ = 0

T = 828 N

Therefore, 828 N is the tension in the cable closer to the painter.

(b) Now, we will calculate the sum about close ends as follows.

$m_{1}g \frac{l_{1}}{2} + m_{2}g \times (1.9) - T \times l_{1}$

$57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (1.9) - T \times 6.5$ = 0

T= 506 N

Therefore, 506 N is the tension in the cable further from the painter.

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3 years ago

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Answer:

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Explanation:

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