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adelina 88 [10]

2 years ago

14

You illuminate the surface of some metal with green laser and observe the photocurrent. If you decrease the intensity of laser,

the stopping potential will

1 answer:

gregori [183]2 years ago

7 0

Remain the same.

Explanation:

The kinetic energy of the ejected electrons, and thus the resulting photocurrent, does not depend on the intensity of the incident radiation. Instead, it depends on the frequency. So since the stopping potential is used to reduce the photocurrent to zero and the photocurrent does not depend on the intensity, the stopping potential remains the same.

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A 20.0 μf capacitor is charged to a potential difference of 850 v. the terminals of the charged capacitor are then connected to

liberstina [14]

(a) $Q = 1.70\times 10^{-2}\;\text{C}$ ;
(b) $V_\text{final} = 5.31\times 10^{2}\;\text{V}$ ;
(c) $E_\text{final} = 4.52\;\text{J}$ ;
(d) $\Delta E = 2.82\;\text{J}$ .

All four values are in 3 sig. fig.

<h3>Explanation</h3>

(a)

$Q = C\cdot V = 20.0\times 10^{-6} \times 850\;\text{V} = 1.70\times 10^{-2}\;\text{J}$ .

(b)

Sum of the final charge on the two capacitors should be the same as the sum of the initial charge. Voltage of the two capacitors should be the same. That is:

$C_1\cdot V_\text{final} +C_2 \cdot V_\text{final} = C_1\cdot V_\text{initial}$ ;

$(C_1+C_2)\cdot V_\text{final} = C_1\cdot V_\text{initial}$ ;

$\displaystyle V_\text{final} = \frac{C_1}{C_1+C_2}\cdot V_\text{initial}\\\phantom{V_\text{final}} = \frac{20.0\;\mu\text{F}}{20.0\;\mu\text{F} + 12.0\;\mu\text{F}} \times 850\;\text{V}\\\phantom{V_\text{final}} =531\;\text{V}$ .

(c)

$\displaystyle E = \frac{1}{2}\cdot C\cdot V^{2}$ .

$\displaystyle E_\text{final} = \frac{1}{2} (C_1 + C_2) \cdot {V_\text{final}}^{2} \\\phantom{E_\text{final}} = \frac{1}{2} \times (20.0\times 10^{-6} + 12.0\times 10^{-6}) \times 531.25\\\phantom{E_\text{final}} = 4.52\;\text{J}$ .

(d)

Initial energy of the system, which is the same as the initial energy in the $20.0\;\mu\text{F}$ capacitor:

$\displaystyle E_\text{initial} = \frac{1}{2} \times 20.0\times 10^{-6} \times 850^{2} = 7.225\;\text{J}$ .

Change in energy:

$\Delta E = 7.225\;\text{J} - 4.516\;\text{J} = 2.70\;\text{J}$ .

4 0

3 years ago

A weather balloon has a volume of 90.0 l when it is released from sea level. What

frozen [14]

The final atmospheric pressure is $5.19\cdot 10^4 Pa$

Explanation:

Assuming that the temperature of the air does not change, we can use Boyle's law, which states that for a gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume. In formula,

$pV=const.$

where

p is the gas pressure

V is the volume

The equation can also be rewritten as

$p_1 V_1 = p_2 V_2$

where in our problem we have:

$p_1= 1.03\cdot 10^5 Pa$ is the initial pressure (the atmospheric pressure at sea level)

$V_1 = 90.0 L$ is the initial volume

$p_2$ is the final pressure

$V_2 = 175.0 L$ is the final volume

Solving the equation for p2, we find the final pressure:

$p_2 = \frac{p_1 V_1}{V_2}=\frac{(1.01\cdot 10^5)(90.0)}{175.0}=5.19\cdot 10^4 Pa$

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3 0

3 years ago

To test the performance of its tires, a car

Rom4ik [11]

The coefficient of static friction is 0.222

Explanation:

In order for the car to remain in circular motion, the frictional force must be able to provide the necessary centripetal force. Therefore, the car will start skidding when the two forces are equal:

$\mu mg=m\frac{v^2}{r}$

where the term on the left is the frictional force, while the term on the right is the centripetal force, and where

$\mu$ is the coefficient of static friction

m is the mass of the car

g is the acceleration of gravity

v is the speed of the car

r is the radius of the track

In this problem, we have:

r = 564 m

v = 35 m/s

$g=9.8 m/s^2$

And re-arranging the equation for $\mu$ , we can find the coefficient of static friction:

$\mu = \frac{v^2}{gr}=\frac{35^2}{(9.8)(564)}=0.222$

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5 0

3 years ago

If a road does not have a bicycle lane, where must a bicyclist ride their bicycle?

DIA [1.3K]

A bicyclist can ride their bicycle still on the road. Bicycle riders be able to take the public ways which has the similar rights and accountability as motorists and are subject to the same guidelines and protocols. The law says that individuals who ride bikes should ride as nearby to the right side of the road as likely excluding under the following conditions: when passing, preparing for a left go, evading risks, if the lane is too constricted to share, or if oncoming a place where a right turn is approved. In a road which has a bike lane the bicyclists roving slower than road traffic must custom the bike way excluding when creating a left turn, passing, evading hazardous settings, or impending a place where a right turn is approved.

5 0

3 years ago

Why cant soy sauce freeze

juin [17]

The salt content in soy

8 0

3 years ago

Read 2 more answers