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Tems11 [23]
2 years ago
13

Which is true?The photoelectric effect occurs only for frequencies below the cutoff frequency, regardless of the intensity.The p

hotoelectric effect occurs only for frequencies above the cutoff frequency, regardless of the intensity.The photoelectric effect occurs only for frequencies below the cutoff frequency and only for intensities above a cutoff intensity.The photoelectric effect occurs only for frequencies above the cutoff frequency and only for intensities above a cutoff intensity.
Physics
2 answers:
Crank2 years ago
7 0

Answer:

The photoelectric effect occurs only for frequencies above the cutoff frequency, regardless of the intensity.

Explanation:

The photoelectric effect occurs when light is shined on metals such as zinc beyond a certain frequency (the threshold frequency), which causes electrons to escape from the zinc. The electrons which are fleeing are called photo electrons.

Therefore photo electric effect is

The photoelectric effect occurs only for frequencies above the cutoff frequency, regardless of the intensity.

SIZIF [17.4K]2 years ago
6 0

Answer:

Explanation:

When a photon of suitable frequency falls on the metal surface, the electrons are ejected from the surface. This minimum frequency required to jest eject the electrons from the metal surface is called cut off frequency.

So, the photoelectric effect occurs only for the frequencies above the cut off frequency regardless of the intensity of photons falling on the metal surface.

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Transcranial magnetic stimulation (TMS) is a noninvasive technique used to stimulate regions of the human brain. A small coil is
attashe74 [19]

Answer:

0.125 volts

Explanation:

The induced emf can be sufficient to stimulate neuronal activity.

One such device generates a magnetic field within the brain that rises from zero to 1.5 T in 120 ms.

We need to find the induced emf within a circle of tissue of radius 1.6 mm and that is perpendicular to the direction of the field. The formula for the induced emf is given by :

\epsilon=-\dfrac{d\phi}{dt}

Where

\phi is magnetic flux

So,

\epsilon=-\dfrac{d(BA)}{dt}\\\\=2\pi r\times \dfrac{dB}{dt}\\\\=2\pi \times 1.6\times 10^{-3}\times \dfrac{1.5-0}{120\times 10^{-3}}\\\\=0.125\ V

So, the induced emf is equal to 0.125 volts.

7 0
3 years ago
A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial sp
Dimas [21]

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m =  0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2mU^{2}

K.E = 1/2 x 0.17 x 6^{2}

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/s^{2}

To calculate the final velocity, let us use third equation of motion.

V^{2} = U^{2} + 2as

V^{2}  = 6^{2} + 2 x 0.3 x 61

V^{2} = 36 + 36

V^{2} = 72

V = \sqrt{72}

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

5 0
2 years ago
a pendulum, 2.0 m in length, is released with a push when the string is at an angle of 25 o from the vertical. if the initial sp
stiks02 [169]

Answer:

1 / 2 m v^2 = L m g (1 - cos θ)

This is the KE due to the pendulum falling from a 25 deg displacement

v^2 = 2 L g (1 - cos 25) = 2 * 2 * 9.8 (1 - .906) = 3.67 m^2/s^2

v = 1.92 m/s      this is the speed due to an initial displacement of 25 deg

Its speed at the bottom would then be

1.92 + 1.2 = 3.12 m/s   since it gains 1.92 m/s from its initial displacement

3 0
1 year ago
Given a second class lever with a distance of 5.00 feet from the fulcrum to the effort and a distance of 33.0 inches from the re
leva [86]

Answer:

The correct answer is C. 45.5 lbs.

Explanation:

In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.

The formula for any problem involving a lever is:

F_ed_e=F_ld_l

Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.

The parameter of the formula that you need is F_l:

F_l=\frac{F_ed_e}{d_l}

The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.

F_l=\frac{25*60}{33}

F_l=45.5 lbs

7 0
3 years ago
Which of the following statements explains how total time spent in the air is affected as a projectile's angle of launch is incr
charle [14.2K]

Answer:

Therefore letter <u>C is the correct answer.</u>

Explanation:

In a projectile motion the total time in the air can be calculated using the following equation:

We analyze the y-component motion.

v_{fy}=v_{iy}-gt

When the final velocity (v(f)) is equal to zero we calculate the upward time and multiplying it by 2 we find the total time in the air. So we will have:

t_{tot}=2\frac{v_{iy}}{g}

t_{tot}=2\frac{v_{i}sin(\theta)}{g}

We can see that the <u>total time is directly proportional to the angle</u>, then when <u>θ increase t increase.</u>

Therefore letter C is the correct answer.

I hope it helps you!  

3 0
3 years ago
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