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adelina 88 [10]
2 years ago
14

You illuminate the surface of some metal with green laser and observe the photocurrent. If you decrease the intensity of laser,

the stopping potential will
Physics
1 answer:
gregori [183]2 years ago
7 0

Remain the same.

Explanation:

The kinetic energy of the ejected electrons, and thus the resulting photocurrent, does not depend on the intensity of the incident radiation. Instead, it depends on the frequency. So since the stopping potential is used to reduce the photocurrent to zero and the photocurrent does not depend on the intensity, the stopping potential remains the same.

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Where could convection currents form? Check all that apply.
KatRina [158]
<h3><u>Answer;</u></h3>
  • In a freshwater lake
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  • In Earth's mantle
<h3><u>Explanation</u>;</h3>
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8 0
3 years ago
Read 2 more answers
. Now assuming Anna's far point was found to be 0.9 m (i.e., her eyes can't focus on any object more than 0.9 m away), what powe
Mama L [17]

Answer:1). Distance of far point x=0.9m

Therefore, since the image is virtual

-f=-x = -0.9m

Power of the concave lenses = 1/f = 1/-0.9

= -1.11D

2 ) near point is 21cm = 0.21m

Power = 4-1/near point

= 4/0.21

= 14.2D.

7 0
4 years ago
a volleyball is hit upward with an initial velocity of 7.5 m/s. calculate the displacement of the volleyball when its final velo
Luden [163]

Answer:

The displacement of the volleyball is 2.62 m

Explanation:

Given;

initial velocity of the volleyball, u = 7.5 m/s

final velocity of the volleyball, v = 2.2 m/s

displacement of the volleyball, d = ?

Apply the following kinematic equation;

v² = u² - 2gd

2gd = u² - v²

d = \frac{u^{2}-v^{2}  }{2g}\\\\d = \frac{7.5^{2}-2.2^{2}  }{2*9.8}\\\\d = 2.62 \ m

Therefore, the displacement of the volleyball is 2.62 m

7 0
3 years ago
The tip of the fan blade is 0.61 m from the center of the fan. The fan turns at a constant speed and completes 2 rotation every
Goryan [66]

Answer:

What is the centripetal acceleration of the tip of the fan blade?

6.0 m/s2

48 m/s2

53 m/s2

96 m/s2

Answer is 96

Explanation:

5 0
3 years ago
Person is lifting a 250 N dumbbell. The weight is 30 cm from the pivot point of the elbow. What force must be exerted five from
qwelly [4]
Refer to the diagram shown below.

The force, F, is applied at 5 cm from the elbow.

For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N

Answer: 1500 N

4 0
3 years ago
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