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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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How much conventional current must you run in a solenoid with radius = 0.05 m and length = 0.39 m to produce a magnetic field in
Ugo [173]

Answer:

Explanation:

radius of the solenoid, r = 0.05 m

length of the solenoid, l = 0.39 m

Magnetic field of the solenoid, B = 2 x 10^-5 T

Number of turns, N = 200

The magnetic field of the solenoid is given by

B=\mu _{0}ni

where, i be the current and n be the number of turns per unit length

n = N / l = 200 / 0.39 = 512.8

2\times 10^{-5}=4 \times 3.14\times 10^{-7}\times 512.8\times i

i = 0.031 A

3 0
3 years ago
Does parallax affect the precision of a measurement that you make
Paul [167]
Yes, parallax affects the precision of a measurement that you make. It introduces an error in the order of the parallax. It will cause the measurement to be different from the real answer. Hope this answers the question. Have a nice day.
8 0
3 years ago
Read 2 more answers
A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help"
Digiron [165]

Answer: a) The cliff is 532.05m high

b) Her speed just before hitting the ground is 102.12 m/s

Explanation: To solve This, I'll use a sketch diagram, attached to this solution,

In 3seconds, the teacher heard the echo of her initial scream back. We can obtain the distance the teacher had fallen at the end of 3 seconds using the equations of motion,

Y1 = ut + 0.5g(t^2)

Since she's falling under the influence of gravity, her initial velocity, u = 0m/s, g = 9.8m/s2, t = 3s

Y1, distance she fell through in 3 seconds = 0.5×9.8(3^2) = 44.1m

Let the total height of the cliff be (44.1 + x); where is the remaining height of cliff that the teacher will fall through.

Using the equations of motion again, we can obtain distance travelled by the sound waves in 3s. sound waves travel with a constant speed of 340m/s, no acceleration,

Y2 = ut + 0.5g(t^2) where g = 0, u = 340m/s, t = 3seconds

Y2 = 340 × 3 = 1020m

But in 3 secs, the sound waves would have travelled through the total height of the cliff (44.1 + x) and back to the teacher's current height, x. That is, 1020 = 44.1 + x + x

x = 487.95m

So, total height of cliff = 44.1 + 487.95 = 532.05m

b) the speed of the teacher just before she hits the ground.

Using the equations of motion again,

(V^2) = (U^2) + 2gs

Where v is the final velocity to be calculated

U is the initial velocity = 0m/s

g is acceleration due to gravity = 9.8m/s2

S is the total height she fell through, that is, the height of the cliff = 532.05m

(V^2) = 0 + 2×9.8×532.05 = 10428.18

V = √(10428.18) = 102.12m/s

QED!

4 0
2 years ago
PLEASE HELP ME ASAP PLEASEEEEE
AleksAgata [21]
I believe its the law of inertia
6 0
3 years ago
A friend tells you that a lunar eclipse will take place the following week, and invites you to join him to observe the eclipse t
WARRIOR [948]

Answer:

y = 80.2 mille

Explanation:

The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening

          θ = 1.22 λ/ d

in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m

         θ = 1.22 550 10⁻⁹ / 0.002

         θ = 3.355 10⁻⁴ rad

Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi

       tan θ = y / L

       y = L tan θ

       y = 2,389 10⁵ tan 3,355 10⁻⁴

       y = 8.02 10¹ mi

       y = 80.2 mille

This is the smallest size of an object seen directly by the eye

5 0
3 years ago
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