(a) A 2.00-µF capacitor is connected to a 18.0-V battery. How much energy is stored in the capacitor?

Answers

Answer 1
Answer:

Answer:

1.8 x 10⁻⁵J

Explanation:

The energy (E) stored in a capacitor of capacitance, C,  when a voltage, V, is supplied is given by;

E = (1)/(2) x C x V²              -------------------(i)

Now, from the question;

C = 2.00μF = 2.00 x 10⁻⁶F

V = 18.0V

Substitute these values into equation (i) as follows;

E = (1)/(2) x 2.00 x 10⁻⁶ x 18.0

E = 1.8 x 10⁻⁵J

Therefore, the quantity of energy stored in the capacitor is 1.8 x 10⁻⁵J

Answer 2
Answer:

Answer:

1.8 x 10⁻⁵J.

Explanation:

E = 1/2 x C x V²

Where,

E = energy stored in a capacitor

C = capaciitance

V = Voltage

From the question, given:

C = 2.00μF

= 2.00 x 10⁻⁶F

V = 18.0V

E = 1/2 x 2.00 x 10⁻⁶ x 18.0

= 1.8 x 10⁻⁵J.


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Solve for x
–30 = 5(x + 1)

Answers

Answer:

-30=5(x+1) is -7

Explanation:

distribute flip subtract 5 from both sides divide both sides by 5

-30=(5)(x)+(5)(1) (Distribute)
-30=5x+5
Flip equation
5x+5=-30
Subtract 5 from both sides
5x+5-5=-30-5
5x=-35
Divide both sides by 5
X=-7

You are exiting a highway and need to slow down on the off-ramp in order to make the curve. It is rainy and the coefficient of static friction between your tires and the road is only 0.4. If the radius of the off-ramp curve is 36 m, then to what speed do you need to slow down the car in order to make the curve without sliding?

Answers

Answer:

11.87m/s

Explanation:

To solve this problem it is necessary to apply the concepts related to frictional force and centripetal force.

The frictional force of an object is given by the equation

F_r = \mu N

Where,

\mu =Friction Coefficient

N = Normal Force, given also as mass for acceleration gravity

In the other hand we have that centripetal force is given by,

F_c= (mv^2)/(R)

The force experienced to stay on the road through friction is equal to that of the centripetal force, therefore

F_r = F_c

\mu mg = (mv^2)/(R)

Re-arrange to find the velocity,

V = √(R\mu g)

V = √(36*0.4*9.8)

V = 11.87m/s

Therefore the speed that it is necessaty to slow down the car in order to make the curve without sliding is 11.87m/s

Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit. The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm. What is the width of the slit?

Answers

Answer:

The width of the slit is 0.167 mm

Explanation:

Wavelength of light, \lambda=608\ nm=608* 10^(-9)\ m

Distance from screen to slit, D = 88.5 cm = 0.885 m

The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m

We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :

y=(mD\lambda)/(a)

where

a = width of the slit

a=(mD\lambda)/(y)

a=(5* 0.885\ m* 608* 10^(-9)\ m)/(0.0161\ m)

a = 0.000167 m

a=1.67* 10^(-4)\ m

a = 0.167 mm

So, the width of the slit is 0.167 mm. Hence, this is the required solution.

An isotropic point source emits light at wavelength 500 nm, at the rate of 185 W. A light detector is positioned 380 m from the source. What is the maximum rate ∂B/∂t at which the magnetic component of the light changes with time at the detector's location?

Answers

Answer:

(dB)/(dt) = 3.49 *10^(6) \ \ T/s

Explanation:

Given that

An isotropic point source emits light at a wavelength \lambda = 500 nm

Power = 185 W

Radius = 380 m

Let's first calculate the The intensity  of the wave , which is = (Power )/(Area)

= (Power)/(4 \pi r ^2)

=  (185 \ W)/( 4 \pi (380)^2)

= 1.0195*10^(-4) \ W/m^2

Now;

The amplitude of the magnetic field is calculated afterwards by using poynting vector

i.e

I = ((c)/(2 \mu_0 ))B_(max^2)

B_(max^2) =  ((2 \mu_0   I)/( c))

B_(max^2) =  ((2 *4 \pi *10^(-7)*1.0195*10^(-4))/( 3*10^8))

B_(max^2) = 8.5409*10^(-19)

B_(max) = \sqrt {8.5409*10^(-19)}

B_(max) = 9.242*10^(-10)

The magnetic field wave equation can now be expressed as;

B = B_(max) sin (kx - \omega t)

Taking the differentiation

(dB)/(dt)= - \omega B_(max) \ cos ( kx - \omega t)

The maximum value ;

(dB)/(dt) = \omega B _(max)

where ;

\omega = 2 \pi f\n\omega = (2 \pi c)/(\lambda)

then

(dB)/(dt) = (2 \pi c)/(\lambda) B _(max)

(dB)/(dt) = (2 \pi 3*10^8*9.242*10^(-10))/(500*10^(-9))

(dB)/(dt) = 3484751.917

(dB)/(dt) = 3.49 *10^(6) \ \ T/s

The maximum rate(∂B/∂t) at which the magnetic component of the light changes with time at the detector's location is approximately6.8 x 10^9 Tesla per second (T/s).

To find the maximum rate at which the magnetic component of the light changes with time at the detector's location, you can use the formula for the rate of change of magnetic field due to an electromagnetic wave. The formula is given by:

∂B/∂t = (2π / λ) * E * c

Where:

∂B/∂tis the rate of change of the magnetic field.

λ is the wavelength of the light.

E is the electric field strength.

c is the speed of light in a vacuum, approximately3 x 10^8 m/s.

You have the wavelength (λ) as 500 nm, which is 500 x 10^-9 meters, and the electric field strength (E) can be calculated using the power (P) and the distance (r) from the source. The power emitted by the source is 185 W, and the distance from the source to the detector is 380 m.

First, calculate the electric field strength (E):

E = sqrt(P / (2π * r^2))

E = sqrt(185 W / (2π * (380 m)^2))

E = sqrt(185 W / (2π * 144400 m^2))

E ≈ 6.325 x 10^-5 N/C

Now, you can calculate the rate of change of the magnetic field:

∂B/∂t = (2π / λ) * E * c

∂B/∂t = (2π / (500 x 10^-9 m)) * (6.325 x 10^-5 N/C) * (3 x 10^8 m/s)

∂B/∂t ≈ (3.77 x 10^15 Hz) * (6.325 x 10^-5 N/C) * (3 x 10^8 m/s)

∂B/∂t ≈ 6.8 x 10^9 T/s

So, the maximum rate at which the magnetic component of the light changes with time at the detector's location is approximately6.8 x 10^9Tesla per second (T/s).

Learn more about magnetic component of the light here:

brainly.com/question/33261927

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A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of 3600 rpm. When the power is turned on, the unit reaches its rated speed in 5 s, and when the power is turned off, the unit coasts to rest in 70 s. Assuming uniformly accelerated motion, determine the number of revolutions that the motor executes (a) in reaching its rated speed, (b) in coasting to rest.

Answers

Answer:

(a) θ1 = 942.5rad, (b) θ2 = 13195 rad

Explanation:

(a) Given

ωo = 0 rad/s

ω = 3600rev/min = 3600×2(pi)/60 rad/s

ω = 377rad/s

t1 = 5s

θ1 = (ω + ωo)t/2

θ1 = (377 +0)×5/2

θ1 = 942.5 rads

(b) ωo = 377rad/s

ω = 0 rad/s

t2 = 70s

θ2 = (ω + ωo)t/2

θ2 = (377 +0)×70/2

θ2 = 13195 rad

A CO2 gun shoots a 0.2 gram round pellet (bb) at 2800 ft/sec, and as the bb leaves the gun it gets charged by friction . If Earths magnetic field points South to North at an intensity of 20 uT, and the bb is shot W->E. Find the charge the bb would need to stay level by balancing out the force of gravity.

Answers

Answer:

So it would need a charge of 0.115C for a upward force to act and cancel the force of gravity.

Explanation:

Fb = Fg

so

qvb= mg ⇒ q = mg/vB = 0.2 *10∧-3 * 9.8/853.44 * 20 * 10∧-6

                                                       = 0.115C

note:2800ft/sec = 853.44m/s

So it would need a charge of 0.115C for a upward force to act and cancel  the force of gravity.