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goblinko [34]
3 years ago
12

Monochromatic light of a given wavelength is incident on a metal surface. However, no photoelectrons are emitted. If electrons a

re to be ejected from the surface, how should the incident light be adjusted?
O Light of a shorter wavelength should be used.
O The intensity of the light should be increased.
O The intensity of the light should be decreased.
O Light of a longer wavelength should be used.
Physics
1 answer:
mrs_skeptik [129]3 years ago
4 0

Answer:

Light of a shorter wavelength should be used.

Explanation:

This is studied in the phenomenon called photoelectric effect, in which light is able to release electrons from a metal, said electrons are called photoelectrons .

The experiments that have been carried out show that <u>increasing  or decreasing the intensity of the light will not cause the photoelectrons to be emitted</u>, what will cause the photoelectrons to be emitted is to increase the frequency of the incident light.

And a higher frequency corresponds to a shorter wavelength according to the equation:

f=\frac{c}{\lambda}

(where f is frequency, c the speed of light, and \lambda the wavelength)

So the answer is that the wavelength of the light must be shortened to cause the emission of electrones.

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A car starting from rest has a constant acceleration of 4 meters per second per second. How fast will it be going in 5 seconds?
irinina [24]

Answer:

18 m

Explanation:

Given : vo = 0 m/s ; t = 3 s; a = 4 m/s^2 ; d = ? m ; average velocity = ? m/s ; fonal velocity = ? m/s

solving for the final velocity, v

v = a * t

v = 4 m/s^2 * 3 s

v = 12 m / s

Solving for the average velocity. avg v

avg v = (vo + v) / 2

avg v = (0 m / s + 12 m/s) / 2

avg v = 6 m / s

Solving for the distance traveled after 3 s

d = avg v * t

d = 6 m / s * 3 s

d = 18 meters

In the first 3s the car travels 18 meters.

4 0
2 years ago
Define moment of momentum. at which condition is it's magnitude zero?​
ololo11 [35]

Let's start with the concept of momentum. What is it? Linear momentum in physics is mathematically written as a product of mass and velocity of an object. Now let us suppose a body of mass m is moving in an inertial frame of reference with velocity v. Consider the fact that no external force is acting on the system. The momentum of this body is given by mv, where m is the mass and v is its velocity. In case of simple real world problems not delving into the realms of relativity, mass is a conserved quantity and it cannot be zero. Hence the velocity of the body must be zero and hence the momentum.

However, photons are considered to have a rest mass zero.

However note the point carefully "rest mass". A body in motion cannot have mass to be zero.

<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em><em> ❤️</em>

7 0
2 years ago
The average distance of the planet mercury from the sun is 0.39 times the average distance of the earth from the sun. How long i
Stels [109]

Answer:

T_1=0.24y

Explanation:

Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:

(\frac{T_1}{T_2})^2=(\frac{D_1}{D_2})^3

Where T_1 and T_2 are the orbital periods of Mercury and Earth respectively. We have D_1=0.39D_2 and T_2=1y. Replacing this and solving for

T_1^2=T_2^2(\frac{D_1}{D_2})^3\\T_1^2=(1y)^2(\frac{0.39D_2}{D_2})^3\\T_1^2=1y^2(0.39)^3\\T_1^2=0.059319y^2\\T_1=0.24y

5 0
3 years ago
A traveler pulls on a suitcase strap at an angle 36° above the horizontal. If of work are done by the strap while moving the sui
ollegr [7]

Answer:

The tension in the strap is 74.82 N.

Explanation:

Given that,

Angle between the horizontal and the suitcase is 36 degrees.

The distance traveled by the suitcase is 15 meters.

Let the work done by the suitcase is 908 J. We know that the work done in the vector form is given by :

W=Fd\ \cos\theta\\\\908=F\times 15\times cos(36)\\\\F=74.82\ N

So, the tension in the strap is 74.82 N. Hence, this is the required solution.

6 0
3 years ago
Use the conditions provided in the previous problem. Atmosphere statically stable at the base of the mountain, where pressure =
Ann [662]

Answer:

change in relative vorticity  0.0590

Explanation:

Given data

pressure = 1000 hPa

temperature lapse rate q1 = 3.1◦C  per 50 hPa

pressure = 850 hPa

temperature lapse rate q2= -0.61◦C per 50 hPa

to find out

change in relative vorticity

solution

we will apply here formula that is

N = (g /  potential temperature ) × (potential vertical temperature) × exp^1/2    ............................1

here we know g = 9.8 m/s

and q1 = potential temperature=3.3 degree celsius

potential vertical temperature gradient = 3.1 - 0.61  / 1000 -850

potential vertical temperature gradient = 0.0166 degree celsius/hpa

so

N = 9.8 / 2.75 × 0.0166 × exp^1/2

N = 0.0590

8 0
3 years ago
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