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2 years ago

15

Light with a photon energy of 3 ev impinges on the surface of a material with work function 0.6 ev to eject electrons. what is t

he kinetic energy of the outgoing electrons

1 answer:

DENIUS [597]2 years ago

5 0

Answer:

KE = incident energy - work function

The maximum KE will be (3 - .6) eV = 2.4 eV

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320 joules of work were done by a force of 80 newtons. Over what

Novay_Z [31]

Answer:

$4\:\mathrm{m}$

Explanation:

The formula for work is given as $W=F\Delta x$ where $F$ is force applied and $\Delta x$ is displacement.

Rearranging this equation, we have:

$\Delta x=\frac{W}{F}$ .

Plugging in given values, we get:

$\Delta x=\frac{W}{F}=\frac{320}{80}=\fbox{$4\:\mathrm{m}$}$ .

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3 years ago

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A transformer has an input of 24 V and an output of 36 V. If the input is changed to 12 V what will the output be?

mote1985 [20]

ANSWER:

18 V

STEP-BY-STEP EXPLANATION:

Given:

Initial condition (1):

input voltage = Vi1 = 24 V

output voltage = Vo1 = 36 V

Final condition (2):

input voltage = Vi2 = 12 V

We can calculate the output voltage of the final condition as follows:

$\begin{gathered} \frac{V_{i1}}{V_{o1}}=\frac{V_{\imaginaryI2}}{V_{o2}} \\ \\ \text{ We replacing:} \\ \\ \frac{24}{36}=\frac{12}{V_{o2}} \\ \\ V_{o2}=\frac{12}{24}\cdot36 \\ \\ V_{o2}=18\text{ V} \end{gathered}$

Therefore, in the final condition, the output voltage will be 18 volts.

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2 years ago

1. A 1.30 kg ball strikes a wall with a velocity of -10.5 m/s. The ball bounces off with a velocity of 6.50 m/s. If the ball is

bekas [8.4K]

Let F be the magnitude of the force. The impulse of this force while the ball is in contact with the wall is

Ft = F (0.0210 s)

and this impulse is equal to the change in the ball's momentum,

m ∆v = (1.30 kg) (6.50 m/s - (-10.5 m/s)) = (1.30 kg) (17.0 m/s)

Solve for F :

F (0.0210 s) = (1.30 kg) (17.0 m/s)

F = (1.30 kg) (17.0 m/s) / (0.0210 s)

F ≈ 1050 N

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_______________________//__/1 amu

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3 years ago

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kakasveta [241]

Answer:

A

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