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2 years ago

In the photoelectric effect, the greater the frequency of the illuminating light, the greater the:_______

Physics

1 answer:

TEA [102]2 years ago

8 0

Answer:

B. Maximum velocity of ejected electrons.

Explanation:

The ejection of electrons form a metal surface when the metal surface is exposed to a monochromatic electromagnetic wave of sufficiently short wavelength or higher frequency (or equivalently, above a threshold frequency), which leads to the enough energy of the wave to incident and get absorbed to the exposed surface emits electrons. This phenomenon is known as the photoelectric effect or photo-emission.

The minimum amount of energy required by a metal surface to eject an electron from its surface is called work function of metal surface.

The electrons thus emitted are called photo-electrons.

The current produced as a result is called photo electricity.

Energy of photon is given by:

$E=h.\nu$

where:

h = Planck's constant

$\nu=$ frequency of the incident radiation.

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After an arrow is shot, is the force unbalanced or balanced? BRAINLY.

Reil [10]

Answer:

The force is unbalanced

Explanation:

After an arrow is shot, the force acting on the arrow is unbalanced. The resulting net force gives the arrow an initial acceleration which wanes with time and the body is brought to rest.

The net force acting on an arrow is not zero and this indicates that the forces acting on the arrow is unbalanced.

If the force is balanced, the arrow is expect to continue in uniform motion but that is not the case as air resistance has massive impact on this body.

7 0

1 year ago

A runner generates 1260 W of thermal energy. If this heat has to be removed only by evaporation, how much water does this runner

sergey [27]

Answer:0.502kg

Explanation:

F4om the relation

Power x time = mass x latent heat of vapourization

P.t=ML

1260 * 15 *60 = M * 22.6 * 10^5

M= 1134000/(22.6 *10^5)

M=0.502kg=502g

3 0

2 years ago

An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t

gtnhenbr [62]

1. $tan \theta = \frac{v^2}{Rg}$

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

$N cos \theta = mg$ (1)

$N sin \theta = m \frac{v^2}{R}$ (2)

where in the second equation, the term $m\frac{v^2}{R}$ represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the following expression:

$tan \theta = \frac{v^2}{Rg}$

2. $F=\frac{m}{R}(w^2-v^2)$

In this second situation, the cars moves around the track at a speed

$w>v$

This means that the centripetal force term

$m\frac{v^2}{R}$

is now larger than before, and therefore, the horizontal component of the normal reaction, $N sin \theta$ , is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

$N sin \theta + F = m\frac{w^2}{R}$

And re-arranging for F,

$F=m\frac{w^2}{R}-N sin \theta$ (3)

But from eq.(2) in the previous part we know that

$N sin \theta = m \frac{v^2}{R}$

So, susbtituting into eq.(3),

$F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)$

4 0

2 years ago

Assume that segment r exerts a force of magnitude t on segment l. what is the magnitude flr of the force exerted on segment r by

mrs_skeptik [129]

If we are talking on the force being exerted by a segment of a rope of lenght R on the right on a point M which is being also pulled from the Left by a segment of rope R as shown in the figure attached. Then we invoke Newton's Third Law:
"Any force exerted by an object (in this case a segment of the rope) also suffers a equal and opposite force".
If we pick $T_R=T$ whis is the tension exerted by the right segment then the left segment will also exert an equal and opposite force so we have that $T_L=-T$