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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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Help me please Am I correct???​
trasher [3.6K]

Answer:

They both experienced the same force as they weigh the same amount

Newtonian physics

Explanation:

5 0
2 years ago
A spherical shell contains three charged objects. The first and second objects have a charge of − 14.0 nC and 33.0 nC , respecti
matrenka [14]

Explanation:

Formula depicting relation between total flux and total charge Q is as follows.

              \phi  = \frac{Q}{\epsilon_{o}}    (Gauss's Law)

Putting the given values into the above formula as follows.

            Q = \phi \times \epsilon_{o}

                = -953 Nm^{2}/C \times 8.854 \times 10^{-12}

                = -8.4 \times 10^{-9} C

                = -8.4 nC

Therefore, when the unknown charge is q  then,

         -14.0 nC + 33.0 nC + q = -8.4 nC

               q = -27.4 nC

Thus, we can conclude that charge on the third object is -27.4 nC.

7 0
3 years ago
A truck moves 70 m east, then moves 120 m west, and finally moves east again a distance of 90 m. If east is chosen as the positi
asambeis [7]

Answer:

140m east

Explanation:

If East is positive then lets rephrase the problem into integers

A truck moves +70 m, then moves -120m, and finally moves +90m.

So totally Displacement = +70-120+90= +140m

Since east is positive, the trucks resultant displacement is 140 m east of origin

6 0
2 years ago
Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm
Burka [1]

Answer:

The atmospheric pressure is 0.97622\times10^{5}\ Pa.

Explanation:

Given that,

Atmospheric pressure P_{atm}= 1.013\times10^{5}\ Pa    

drop height h'= 27.1 mm

Density of mercury \rho= 13.59 g/cm^3

We need to calculate the height

Using formula of pressure

p = \rho g h

Put the value into the formula

1.013\times10^{5}=13.59\times10^{3}\times9.8\times h

h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}

h=0.76\ m

We need to calculate the new height

h''=h - h'

h''=0.76-27.1\times10^{-3}

h''=0.76-0.027

h''=0.733\ m

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

P=\rho g h

Put the value into the formula

P= 13.59\times10^{3}\times9.8\times0.733

P=0.97622\times10^{5}\ Pa

Hence, The atmospheric pressure is 0.97622\times10^{5}\ Pa.

7 0
3 years ago
How can we protect space shuttles or astronauts from space radiation in the absence of the atmospheric layer?
Arisa [49]
Earth's protective magnetic bubble, called the magnetosphere, deflects most solar particles, but in the absence of atmospheric layer, Polyethylene is a good shielding material because it has high hydrogen content, and hydrogen atoms are good at absorbing and dispersing radiation.

The Earth’s atmospheres are kept in place by gravity. The air near the ground is pulled on by gravity and compressed by the air higher in the sky. This causes the air near the ground to be denser and creating different layers with different qualities in which are the atmosphere.

Hope that helps ^^
7 0
3 years ago
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