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const2013 [10]
3 years ago
15

What is the total flux φ that now passes through the cylindrical surface? enter a positive number if the net flux leaves the cyl

inder and a negative number if the net flux enters the cylnder. what is ex(p), the value of the x-component of the electric field produced by by the line of charge at point p which is located at (x,y) = (a,0), where a = 9.2 cm?
Physics
1 answer:
trasher [3.6K]3 years ago
8 0

Net flux through the cylindrical surface is given as

\phi = \frac{q}{epsilon_0}

here q = enclosed charge in the surface

so here in order to find the value of q

q = \lambda* L

so now we have

\phi = \frac{\lambda * L}{\epsilon_0}

so this is the total flux

now by Gauss's law we can find the electric field

\int E.dA = \phi

\int E.dA = \frac{\lambda * L}{\epsilon_0}

E* 2\pi rL = \frac{\lambda * L}{epsilon_0}

E = \frac{\lambda}{2\pi \epsilon_0 r}

<em>by above expression we can find the electric field at required position</em>

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Answer:

The answer to the question is;

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Explanation:

To solve the question we first resolve the units of the given quantities as follows

The gyroscopes spin at 19,200 rpm that is 19,200 revolutions per minute

1 revolution = 2π rad and

1 minute = 60 seconds

Therefore 19,200 revolutions per minute = 2π×19,200÷60 rad/s

= 2010.619 rad/s

The angle of precess is given as 1.0×10⁻⁶ °. We convert the angle to radians as follows

360 ° = 2π radians

1 ° = \frac{\pi }{180} radians and

1.0×10⁻⁶ ° =  \frac{\pi }{180} radians × 1.0×10⁻⁶ ° = 1.745×10⁻⁸ rad

To find the torque we note that the torque is given by

Precession angular speed × The moment of inertia × angular velocity

The precession angular speed is given by \frac{Precession. Angle}{time}

The precession angle was determined in rad as 1.745×10⁻⁸ rad

The precession time is 5 hours which is equal to 5×60×60 = 18000 s

Therefore the precession velocity = \frac{1.745*10^{-8} rad}{18000 s} =  9.696×10⁻¹³ rad/s

The moment of inertia is given by

Formula for the moment of inertia of a thin walled cylinder I = m·r²

Where:

r = Radius of the gyroscope = Diameter/2 = 5.0 cm/2 = 2.5 cm = ‪0.025‬ m

m = Mass of each gyroscopes = 2.0 kg

Therefore I = m·r² = 2.0 kg × (0.025‬ m)² = 0.00125 kg·m²

Torque, τ = Ω·I·ω

Where:

Ω = Precession velocity = 9.696×10⁻¹³ rad/s

I = Moment of inertia = 0.00125 kg·m²

ω = Angular speed = 2010.619 rad/s

τ = 9.696×10⁻¹³ rad/s × 0.00125 kg·m² × 2010.619 rad/s =

2.44 ×10⁻¹² kg·m²/ s² =   2.44 ×10⁻¹² N·m  

The required torque is 2.44 ×10⁻¹² N·m.

7 0
3 years ago
A series AC circuit contains a resistor, an inductor of 150 mH, a capacitor of 5.00 mF, and a generator with DVmax 5 240 V opera
yanalaym [24]

Given:

Inductance, L = 150 mH

Capacitance, C = 5.00 mF

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frequency, f = 50Hz

I_{max} = 100 mA

Solution:

To calculate the parameters of the given circuit series RLC circuit:

angular frequency, \omega =  2\pi f = 2\pi \times50 = 100\pi

a). Inductive reactance,  X_{L} is given by:

\X_{L} = \omega L = 100\pi \times 150\times 10^{-3} = 47.12\Omega

X_{L} = 47.12\Omega 

b). The capacitive reactance,  X_{C} is given by:

\X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50\times 5.00\times 10^{-3}} = 0.636\Omega

X_{C} = 0.636\Omega

c). Impedance, Z = \frac{\Delta V_{max}}{I_{max}} = \frac{240}{100\times 10^{-3}} = 2400\Omega

Z = 2400\Omega

d). Resistance, R is given by:

Z = \sqrt {R^{2} + (X_{L} - X_{C})}

2400^{2} = R^{2} + (47.12 - 0.636)^{2}

R = \sqrt {5757839.238}

R = 2399.5\Omega

e). Phase angle between current and the generator voltage is given by:

tan\phi = \frac{X_{L} - X_{C}}{R}

\phi =tan^{-1}( \frac{X_{L} - X_{C}}{R})

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\phi = 1.11^{\circ}

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Explanation:

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NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supp
IRINA_888 [86]

Complete question :

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Maximum weight of supplies as measured on EARTH :

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3 years ago
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The book that weighs less
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