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3 years ago

For a cylindrical capacitor, the capacitance does not depend on which of the following values?

Physics

1 answer:

IgorC [24]3 years ago

5 0

Answer:

Capacitance of cylindrical capacitor does not depends on the amount of charge on the conductors

Explanation:

Consider a cylindrical capacitor of length L, inner radius R₁ and outer radius R₂, permitivity ε₀ constant then capacitance of cylindrical capacitor is given by:

$C=\frac{2\pi \epsilon_{o}L}{ln\frac{R_{2} }{R_{1}} }$

From this equation it is clear that capacitance of cylindrical capacitor is independent of the amount of charge on the conductors where as directly proportional permitivity constant and length of cylinder where as inversely proportional to natural log of ratio of R₂ and R₁

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andreev551 [17]

When the object is big enough to contract itself into a ball.

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3 years ago

A national college researcher reported that 67%of students who graduated from high school in 2012 enrolled in college. Twenty-ni

dsp73

Answer:

Explanation:

Theorem of Binomial Distribution will apply here.

n = 29 , p = .67 , q = 0.33

mean = np = 29 x .67 = 19.43

Standard Deviation = √npq

= √29 x .67 x .33

= √6.4

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3 years ago

To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d

Nimfa-mama [501]

Complete Question

The compete question is shown on the first uploaded question

Answer:

The speed is $v = 350 \ m/s$

Explanation:

From the question we are told that

The distance of separation is d = 4.00 m

The distance of the listener to the center between the speakers is I = 5.00 m

The change in the distance of the speaker is by $k = 60 cm = 0.6 \ m$

The frequency of both speakers is $f = 700 \ Hz$

Generally the distance of the listener to the first speaker is mathematically represented as

$L_1 = \sqrt{l^2 + [\frac{d}{2} ]^2}$

$L_1 = \sqrt{5^2 + [\frac{4}{2} ]^2}$

$L_1 = 5.39 \ m$

Generally the distance of the listener to second speaker at its new position is

$L_2 = \sqrt{l^2 + [\frac{d}{2} ]^2 + k}$

$L_2 = \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}$

$L_2 = 5.64 \ m$

Generally the path difference between the speakers is mathematically represented as

$pD = L_2 - L_1 = \frac{n * \lambda}{2}$

Here $\lambda$ is the wavelength which is mathematically represented as

$\lambda = \frac{v}{f}$

=> $L_2 - L_1 = \frac{n * \frac{v}{f}}{2}$

=> $L_2 - L_1 = \frac{n * v}{2f}$

Here n is the order of the maxima with value of n = 1 this because we are considering two adjacent waves

=> $5.64 - 5.39 = \frac{1 * v}{2*700}$

=> $v = 350 \ m/s$

7 0

3 years ago

Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the

joja [24]

As it is given that the air bag deploy in time

$t = 10 ms = 0.010 s$

total distance moved by the front face of the bag

$d = 20 cm = 0.20 m$

Now we will use kinematics to find the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$0.20 = 0 + \frac{1}{2}a*0.010^2$

$0.20 = 5 * 10^{-5}* a$

$a = 4000 m/s^2$

now as we know that

$g = 10 m/s^2$

so we have

$a = 400g$

so the acceleration is 400g for the front surface of balloon

3 0

3 years ago

a 360 mile trip began on a greeway in a car traveling at 62 mph. Once the road became a 2 lane highway, the car slowed to 54 mph

Daniel [21]

Answer:

Time spent on the greenway road = 4.5 hours

Time spent on the 2 lane road = 1.5 hours

Explanation:

The distance of the trip is 360 miles and the initial speed of the car is 62 miles/hr and after the road became 2 lane highway the car slowed to 54 miles/hr.

Let us divide the trip into two

Greenway

speed = distance/time

speed = 62 mph

time = a

distance = speed × time

distance = 62a

2 lane highway

speed = distance/time

speed = 54 mph

time = b

distance = speed × time

distance = 54b

Total distance

62a + 54b = 360......................(i)

Total time

a + b = 6..............................(ii)

a = 6-b

insert a in equation (i)

62(6-b) + 54b = 360

372 - 62b + 54b = 360

-8b = 360-372

-8b = - 12

b = 12/8

b = 1.5

from equation (ii)

a + 1.5 = 6

a = 6 - 1.5

a = 4.5

8 0

3 years ago