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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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Lilit [14]
Bruh huh.............
6 0
2 years ago
What resistance would produce a current of 200A with a potiential difference of 200V?
viktelen [127]
Aw, I hate physics, is this on Apex?

Resistance can be calculated with the information given in the question.
Equation for Resistance: R = V/I
V (voltage) = 200 Volts
I (current) = 200 Amps

So 200 divided by 200 = freaking 1

Answer: R = 1 (ohms)

Hope this Helps!

5 0
3 years ago
A toy cannon uses a spring to project a 5.24-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force
salantis [7]

Answer:

Speed will be equal to 1.40 m/sec

Explanation:

Mass of the rubber ball m = 5.24 kg = 0.00524 kg

Spring is compressed by 5.01 cm

So x = 5.01 cm = 0.0501 m

Spring constant k = 8.08 N/m

Frictional force f = 0.031 N

Distance moved by ball d = 15.8 cm = 0.158 m

Energy gained by spring

KE=\frac{1}{2}kx^2=\frac{1}{2}\times 8.08\times 0.0501^2=0.0101J

Energy lost due to friction

W=Fd=0.031\times 0.158=0.0048J

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J

This energy will be kinetic energy

\frac{1}{2}mv^2=0.0052

\frac{1}{2}\times 0.00524\times v^2=0.0052

v = 1.40 m/sec

7 0
3 years ago
The relationship between distance from the sun and orbital period is that as the distance from the sun increases, the orbital pe
iren [92.7K]
Second law is it:))))))))
4 0
2 years ago
A 0.5-kg ball moving at 5 m/s strikes a wall and rebounds in the opposite direction with a speed of 2 m/s. If the impulse occurs
mart [117]

<em>Given that:</em>

                       mass of the ball (m) = 0.5 Kg ,

                    ball strikes the wall (v₁) = 5 m/s ,

rebounds in opposite direction (v₂) = 2 m/s,

                                time duration (t) = 0.01 s,

        <em> Determine the force (F) = ?</em>

We know that from Newton's II law,

                                <em>F = m. a</em>  Newtons  

                                  (velocity acting in opposite direction, so <em>a = ( (v₁ + v₂)/t</em>

                                   = m × (v₁ + v₂)/t

                                   = 0.5 × (5 + 2)/0.01

                                  = 350 N

<em>The force acting up on the ball is 350 N</em>

                                     

6 0
3 years ago
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