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Allisa [31]
3 years ago
13

A simple series circuit consists of a 120 Ω resistor, a 21.0 V battery, a switch, and a 3.50 pF parallel-plate capacitor (initia

lly uncharged) with plates 5.0 mm apart. The switch is closed at t =0s .
Required:
a. After the switch is closed, find the maximum electric flux through the capacitor.
b. After the switch is closed, find the maximum displacement current through the capacitor.
c. Find the electric flux at t =0.50ns.
d. Find the displacement current at t =0.50ns.
Physics
1 answer:
slega [8]3 years ago
8 0

Answer

Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)

Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.

ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)

Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC

30.7pC/εo = 3.47 V∙m <----- C)

ic(0.5ns) = 29.7ma <----- D)

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5 0
3 years ago
An applied force of 20 N is used to accelerate an object to the right across a
vichka [17]

Answer:

F_{norm} = 100 N

F_{net}=10 N

\mu = 0.10

m = 10 kg

a=1.0 m/s^2

Explanation:

To determine the normal force, we just need to analyze the situation along the vertical direction.

The box along the vertical direction is in equilibrium, so the equation of the forces is

F_{norm} - F_{grav} = 0

which means that

F_{norm} = F_{grav} = 100 N

The net force can be determined by looking at the situation along the horizontal direction (since the net force in the vertical direction) is zero. Here we have:

- An applied force of 20 N forward, F_{app} = 20 N

- A frictional force of 10 N backward, F_{frict} = 10 N

So, the net force is

F_{net}=F_{app}-F_{frict}=20-10 = +10 N in the forward direction

The expression for the frictional force is

F_{frict} = \mu F_{norm}

where \mu is the coefficient of friction. Solving for \mu,

\mu = \frac{F_{frict}}{F_{norm}}=\frac{10}{100}=0.10

The force of gravity is given by

F_{grav}=mg

where m is the mass of the object and g=10 m/s^2. Solving for m, we find the mass of the object:

m=\frac{F_{grav}}{g}=\frac{100}{10}=10 kg

Finally, the acceleration can be found by using Newton's second law

F_{net} = ma

where a is the acceleration. Solving for a,

a=\frac{F_{net}}{m}=\frac{10}{10}=1.0 m/s^2

4 0
3 years ago
A particle has an acceleration of 6.14 m/s2 for 0.300 s. at the end of this time the particle's velocity is 9.41 m/s. part a wha
Varvara68 [4.7K]

The initial velocity of particle is 7.568 m/s

Given:

Acceleration = a = 6.14 m/s2

Time = t = 0.300 s.

final velocity = v = 9.41 m/s.

To Find:

initial velocity = u

Solution: Velocity is the prime indicator of the position as well as the rapidity of the object.

v = u + at

u = v - at = 9.41 - 6.14*0.3 = 7.568 m/s

Hence, initial velocity of particle is 7.568 m/s

Learn more about Velocity here:

brainly.com/question/25749514

#SPJ4

4 0
2 years ago
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Triss [41]
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8 0
2 years ago
a girl performed 50j of work lifting a heavy box it took her 5 seconds to lift the box what is her power
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Answer:

10 W

Explanation:

Power is work over time.

P = W / t

P = 50 J / 5 s

P = 10 W

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