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3 years ago

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

Physics

1 answer:

TEA [102]3 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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A ball with a horizontal speed of 1.0m/s rolls off a bench 2.0 m high. (a) how long will the ball take to reach the floor? (b) h

bixtya [17]

The motion of the ball is a composition of two motions:
- on the x (horizontal) axis, it is a uniform motion with initial velocity $v_x = 1.0 m/s$
- on the y (vertical) axis, it is a uniformly accelerated motion with acceleration $g= 9.81 m/s^2$

(a) to solve this part, we just analyze the motion on the vertical axis. The law of motion here is
$y(t) = h - \frac{1}{2} gt^2$
By requiring y(t)=0, we find the time t at which the ball reaches the floor:
$h- \frac{1}{2}gt^2=0$
$t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2\cdot 2.0 m}{9.81 m/s^2} }=0.64 s$

(b) for this part, we can analyze only the motion on the horizontal axis. To find how far the ball will land, we must calculate the distance covered on the x-axis, x(t), when the ball reaches the ground (so, after a time t=0.64 s):
$x(t) = v_x t = (1.0 m/s)(0.64 s)=0.64 m$

4 0

3 years ago

What is the initial vertical velocity?

Marat540 [252]

Answer:

Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m below its starting altitude spends 3.79 s in the air.The initial vertical velocity is the vertical component of the initial velocity: v 0 y = v 0 sin θ 0 = ( 30.0 m / s ) sin 45 ° = 21.2 m / s .

3 0

2 years ago

What is the value of ΔVBA=VB−VA if the charge on the plates is 1.00 x 10-9 C, the area of the plates is 2.00 m2 and the distanc

Murljashka [212]

Answer:

The Value is $V_{AB} = 2.825V$

Explanation:

The explanation is shown on the first uploaded image

5 0

3 years ago

An object with a charge of -3.2 uC and a mass of 1.0Ã—10^(-2) kg experiences an upward electric force, due to a uniform electric

USPshnik [31]

Answer:

The magnitude of the acceleration is equal to 19.6m/s² and the acceleration is directed upwards though the magnitude of the charge has doubled. This is because the electric force is directed upwards and from newton's second law of motion the charge will have acceleration in the same direction as the electric force on the charge.

Explanation:

The detailed solution can be found in the attachment below.

Thank you for reading and I hope this is helpful to you.

6 0

3 years ago

A third baseman makes a throw to first base 40.5 m away. The ball leaves his hand with a speed of 30.0 m/s at a height of 1.4 m

Ugo [173]

Answer:

When the ball goes to first base it will be 4.23 m high.

Explanation:

Horizontal velocity = 30 cos17.3 = 28.64 m/s

Horizontal displacement = 40.5 m

Time

$t=\frac{40.5}{28.64}=1.41s$

Time to reach the goal posts 40.5 m away = 1.41 seconds

Vertical velocity = 30 sin17.3 = 8.92 m/s

Time to reach the goal posts 40.5 m away = 1.41 seconds

Acceleration = -9.81m/s²

Substituting in s = ut + 0.5at²

s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m

Height of throw = 1.4 m

Height traveled by ball = 2.83 m

Total height = 2.83 + 1.4 = 4.23 m

When the ball goes to first base it will be 4.23 m high.

8 0

3 years ago