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mixer [17]
3 years ago
12

A 2.5-cm-diameter parallel-plate capacitor has a 2.2 mm spacing. The electric field strength inside the capacitor is 6.0×104 V/m

. You may want to review (Pages 699 - 702) . Part APart complete What is the potential difference across the capacitor? Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
vladimir2022 [97]3 years ago
3 0

Explanation:

The given data is as follows.

     Separation between two plates, d = 2.2 mm = 2.2 \times 10^{-3} m

     Electric field strength inside the capacitor, E = 6.0 \times 10^{4} V/m

Therefore, we will calculate the potential difference across the capacitor as follows.

              V = Ed

                  = 6.0 \times 10^{4} V/m \times 2.2 \times 10^{-3} m

                  = 1.32 V

Thus, we can conclude that the potential difference across the given capacitor is 1.32 V.

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What is the freezing point of an aqueous solution that boils at 101 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your an
astra-53 [7]

Answer:

-3.63 degree Celsius

Explanation:

We are given that

Boiling point of solution=T_b=101^{\circ} C

Boiling point water=100 degree Celsius

K_f=1.86K/m

K_b=0.512 K/m

\Delta T_b=T-T_0

Where T=Boiling point of solution

T_0=Boiling point of pure solvent

\Delta T_b=101-100=1^{\circ}C

\Delta T_b=k_bm

Using the formula

1=0.512\times m

Molality,m=\frac{1}{0.512} m

\Delta T_f=k_fm

Using the formula

\Delta T_f=\frac{1}{0.512}\times 1.86

\Delta T_f=3.63 C

We know that

\Delta T_f=T_0-T_1

Where T_0 =Freezing point of solvent

T_1= Freezing point of solution

Using the formula

3.63=0-T_1

Freezing point of water=0 degree Celsius

T_1=0-3.63=-3.63 C

Hence, the freezing point of solution=-3.63 degree Celsius

7 0
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How many hour are required to make a 3000 km trip if your average speed is 50 km/h?
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A machine with a 5-N input force and a 25-N output force has a mechanical advantage of?
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On the Moon's surface, lunar astronauts placed a corner reflector, off which a laser beam is periodically reflected. The distanc
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Answer:

Explanation:

The difference in time will be due to travel through atmosphere where speed of light slows down. If t be the thickness of atmosphere and c be the speed of light in space and μ be the refractive index of atmosphere difference in travel time will be as follows .

difference = \frac{2t\mu }{c}-\frac{2t }{c}

=\frac{2t}{c }\left ( 1-\mu  \right )

Now t = 40 x 10³m ,μ = 1.000293 , c = 3 x 10⁸.

difference =\frac{2t\mu }{c}-\frac{2t }{c}

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