Answer:
**Answer:**

**Explanation:**

1.8 x 10⁻⁵J

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

E = 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 = 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:

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.

An advantage of J.J. Thomson's Plum Pudding Model was that it _____. A. was a much less expensive way to study atoms B. simplified the calculations necessary to describe an atomC. clearly explained where electrons were located in an atomD. is much less expensive to bake a plum pudding than to look at an atom

For every increase in mass the gravitational force blank If the total mass increase by effective for the gravitational force

What is the magnitude of a point charge that would create an electric field of 1.18 N/C at points 0.822 m away?

What is matter? explain and give example

An athlete can exercise by making mechanical waves in ropes. What is themedium of these waves?A. EnergyB. The ropeC. The athleteD. Air

For every increase in mass the gravitational force blank If the total mass increase by effective for the gravitational force

What is the magnitude of a point charge that would create an electric field of 1.18 N/C at points 0.822 m away?

What is matter? explain and give example

An athlete can exercise by making mechanical waves in ropes. What is themedium of these waves?A. EnergyB. The ropeC. The athleteD. Air

–30 = 5(x + 1)

**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

-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

**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

Where,

Friction Coefficient

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

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

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

Re-arrange to find the velocity,

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

**Answer:**

The width of the slit is **0.167 mm**

**Explanation:**

Wavelength of light,

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 :**

**where**

**a = width of the slit**

a = 0.000167 m

**a = 0.167 mm**

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

**Answer:**

**Explanation:**

Given that

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

Power = 185 W

Radius = 380 m

Let's first calculate the The intensity of the wave , which is =

=

=

=

Now;

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

i.e

The magnetic field wave equation can now be expressed as;

Taking the differentiation

The maximum value ;

where ;

then

The maximum rate at which the** magnetic component of the light** changes with time at the detector's location is approximately 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:

Where:

is 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, approximately

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):

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

So, the maximum rate at which the magnetic component of the light changes with time at the detector's location is approximatelyTesla per second (T/s).

Learn more about ** magnetic component of the light ** here:

#SPJ6

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

**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.