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Jet001 [13]
3 years ago
13

A laser emits a beam of light whose photons all have the same frequency. When the beam strikes the surface of a metal, photoelec

trons are ejected from the surface. What happens if the laser emits twice the number of photons per second
Physics
1 answer:
seraphim [82]3 years ago
3 0

Answer:

the no. of ejected electrons per second will increase.

Explanation:

In photoelectric effect, when a light is incident on a metal surface it ejects some electrons from the metal surface. The energy of photon of light must be equal to or greater than the work function of that metal. All the extra energy above the work potential appears as the kinetic energy of the ejected electrons. So, greater he energy of photon greater will be the kinetic energy of the ejected electrons.

A single photon interacts with a single electron and ejects it only if its energy is greater than work function. So, the increase in no. of photons per second means an increase in the intensity of laser beam. And greater no. of photons, will interact with greater no. of electrons. So, <u>the no. of ejected electrons per second will increase.</u>

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You are conducting an experiment inside a train car that may move along level rail tracks. A load is hung from the ceiling on a
katrin2010 [14]

Answer:

b

c

e

h

Explanation:

Note that the swing direction was not giving in the question and direction could be sideways (in a turn) or in a track or both

The question show something in common ...acceleration

So let's look at the statements and pick the correct ones

a is false while b is correct as the train is accelerating

c is correct. The train is accelerating even thou the speed could not be ascertained

d is false and not feasible as the train is accelerating

e is true as the train maybe moving at a constant speed in a circle

f is false. This could be constant velocity in a circle. Same as g (false)

h is true. It's accelerating

7 0
3 years ago
A hockey puck wont slide very far on concrete or asphalt, but it will slide for a very long time on ice. Why is this?
Ilya [14]
When hockey players push the puck along the ice it slides causing heat which melts the ice causing the friction against the ice to be less.
7 0
3 years ago
Read 2 more answers
Which statement best describes perigee?
Margarita [4]
A. is meaningless.
B. is perigee.
C. is apogee.
D. is perihelion.
5 0
3 years ago
Read 2 more answers
At the train station, you notice a large horizontal spring at the end of the track where the train comes in. This is a safety de
e-lub [12.9K]

Assume a maximum stopping acceleration of g/2 where g is acceleration due to gravity.

Answer:

2.99 m/s

Explanation:

Stopping distance, s = 3 ft = 0.914 m

final velocity, v = 0

a = g/2 = 4.9 m/s²

Use third equation of motion:

v^2-u^2 = 2as

substitute the values to find the speed of train:

0 -u^2 = 2\times -4.9 \times 0.914 \\u^2=8.96 \\u=2.99 m/s

6 0
3 years ago
A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w
amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁  + m₂u₂ = v(m₁  +  m₂)

10 x 3   + 25 x 0   = v(10  +  25)

30  = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\

the final kinetic energy of the package;

K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J

the fraction of the initial energy lost;

= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71

7 0
3 years ago
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