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jolli1 [7]
3 years ago
6

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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In a mail-sorting facility, a 2.50-kg package slides down an inclined plane that makes an angle of 20.0° with the horizontal. Th
lawyer [7]

Answer:

The coefficient of kinetic friction is 0.382.

Explanation:

Given:

Angle of inclination is, \theta=20.0°

Mass of package is, m=2.50\ kg

Initial speed of package is, u=2.00\ m/s

Final speed of the package at the bottom is, v=0\ m/s

Distance of travel along the incline is, d=12.0\ m

Acceleration due to gravity is, g=9.8\ m/s^2

Let the coefficient of kinetic friction be \mu.

Now, the frictional force will be acting along the incline but in the direction opposite to the direction of motion.

So, the net acceleration acting on the package will be up the incline and is equal to:

a=\mu g\cos\theta-g\sin\theta ----------------- 1

Now, using equation of motion, we have:

v^2-u^2=2ad\\\\0-(2.00)^2=2a(12.0)

Solving for 'a', we get:

-4.00=24.0a\\\\a=-\frac{4}{24}=-\frac{1}{6}\ m/s^2

Now, plug in the value of 'a' in equation (1). This gives,

\mu g\cos\theta-g\sin\theta=\frac{1}{6} ( Neglecting negative sign)

Plug in all the given values and solve for \mu. This gives,

9.8(-sin(20)+\mu cos(20))=\frac{1}{6}\\\\-0.342+\mu\times 0.94=0.017\\\\0.94\mu=0.342+0.017\\\\0.94\mu=0.359\\\\\mu=\frac{0.359}{0.94}=0.382

Therefore, the coefficient of kinetic friction is 0.382.

5 0
3 years ago
A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W
weqwewe [10]

Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

gauge pressure, P_{gauge \ pressure} = 2.95 atm

Atmospheric pressure, P_{atm} = 101325 Pa

To determine the speed of the water, apply Bernoulli's equation;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2

where;

P₁ = P_{gauge \ pressure} + P_{atm \ pressure}

P₂ = P_{atm}

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 =  P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 =  P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 =  P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} +  \rho gz_1 =  \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} +  \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} +  \rho gz_1)}{\rho} }

where;

\rho is the density of seawater = 1030 kg/m³

v_2 = \sqrt{ \frac{2(2.95*101325 \ + \  1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s

Therefore, the water is flowing at the rate of 28.04 m/s.

7 0
3 years ago
A force of 4 kg weight acts on a body of mass 9.8 kg calculate the acceleration
White raven [17]

Answer:

here given is a weight

then force becomes mg

that is F=Mg

=4*9.8

then by using the formula

F=Ma

a=F/M

=4*9.8/9.8

=4

Explanation:

3 0
3 years ago
Read 2 more answers
How does air resistance affect an object’s speed?
tatuchka [14]
When air resistance<span> acts, acceleration during a fall </span>will<span> be less than g because </span>air resistance affects<span> the motion of the falling </span>objects<span> by slowing it down. </span>Air resistance<span> depends on two important factors - the</span>speed<span> of the </span>object<span> and its surface area. Increasing the surface area of an </span>object<span> decreases its </span>speed<span>.</span>
4 0
2 years ago
You can enter compound units that are combinations of other units that are multiplied together. to enter the ⋅ explicitly, type
alexdok [17]

Answer:

30 N \cdot m

Explanation:

The torque applied by a force can be calculated as

\tau = F d sin \theta

where

F is the magnitude of the force

d is the length of the arm

\theta is the angle between the direction of the force and the arm

In this problem, we have

F = 15 N

d = 2.0 m

\theta=90^{\circ}

Substituting into the equation, we find

\tau = (15)(2.0) sin 90^{\circ}=30 N \cdot m

7 0
3 years ago
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