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2 years ago

Which statement describes how the photoelectric effect causes an electric

Physics

2 answers:

alina1380 [7]2 years ago

8 0

Answer:

A. Light acts as particles, causing electrons on the surface it strikes to be destroyed.

Explanation:

The photoelectric effect is a phenomenon that occurs when light shined onto a metal surface causes the ejection of electrons from that metal. It was observed that only certain frequencies of light are able to cause the ejection of electrons.

seraphim [82]2 years ago

7 0

Answer: C

Explanation: I just took the test

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A wave has a frequency of 15,500 Hz and a wavelength of 0.20 m. What is the

Hoochie [10]

Answer:

3100 m/s

Explanation:

The relationship between frequency and wavelength of a wave is given by the wave equation:

$v=f\lambda$

where

v is the speed of the wave

f is its frequency

$\lambda$ is the wavelength

For the wave in this problem,

f = 15,500 Hz

$\lambda=0.20 m$

Therefore, the wave speed is

$v=(15500)(0.20)=3100 m/s$

4 0

3 years ago

12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L

Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

6 0

3 years ago

REMARKS The speed found in part (a) is the same as if the woman fell vertically through a distance of 21.9 m. The result of part

sasho [114]

Answer:

Yes, if the system has friction, the final result is affected by the loss of energy.

Explanation:

The result that you are showing is the conservation of mechanical energy between two points in the upper one, the energy is only potential and the lower one is only kinetic.

In the case of some type of friction, the change in energy between the same points is equal to the work of the friction forces

$W_{fr}$ = ΔEm

$W_{fr}$ = $Em_{f}$ -Em₀

As we can see now there is another quantity and for which the final energy is lower and therefore the final speed would be less than what you found in the case without friction.

$Em_{f}$ = $W_{fr}$ + Em₀

Remember that the work of the rubbing force is negative, let's write the work of the rubbing force explicitly, to make it clearer

½ m v² = -fr d + mgh

v = √(-fr d 2/m + 2 gh)

v = √ (2gh - 2fr d/m)

Now it is clear that there is a decrease in the final body speed.

Consequently, if the system has friction, the final result is affected by the loss of energy.

5 0

2 years ago

The Grand Canyon is 1800 meters deep at its deepest point. A rock is dropped from the rim above this point. Write the height of

snow_tiger [21]

Answer:

Explanation:

Given

height of grand canyon is $h=1800\ m$

Rock is dropped i.e. initial velocity is zero

$u=0$

Using Equation of motion

$h=ut+\frac{1}{2}at^2$

h=height

u=initial velocity

a=acceleration

t=time

here a=acceleration due to gravity

$1800=0+\frac{1}{2}\times 9.8\times t^2$

$t^2=\frac{2\times 1800}{9.8}$

$t=19.166\ s$

3 0

3 years ago

A man with a weight of 100 N is located

valentina_108 [34]

Answer:

Given,

mass of man = 100 N = 10 kg

height = h = 25m

since the man does not move anything with his force, work done by him is zero

work done on the man = gain in potential energy

P.E=mgh

P.E=10×9.8×25

P.E=2.45KJ

Explanation:

so, potential energy gained by man is 2.45 KJ

5 0

2 years ago