If I am given a total capacitence of two capacitors, their capacitence togather is 22 F. What capacitence would the individual capacitors have if they are connected in parallel or connected in series.

Answers

Answer 1
Answer:

Answer:

In parallel combination, the capacity of each capacitor is 11 F.

In series combination, the capacity of each capacitor is 44 F.

Explanation:

Let there are two capacitors each of capacitance C.

When they are connected in parallel:

In parallel combination, the effective capacitance is Cp.

Cp = C1 + C2 = C + C

22 = 2 C

C = 11 F

When they are connected in series:

In parallel combination, the effective capacitance is Cs.

1 / Cs = 1 / C1 + 1 / C2 = 1 / C + 1 / C = 2 / C

1 / 22 = 2 / C

C = 44 F


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If a material has an index of refraction of 1.61, Determine the speed of light through this medium
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2. An electrical heater 200mm long and 15mm in diameter is inserted into a drilled hole normal to the surface of a large block of material having a thermal conductivity of 5W/m·K. Estimate the temperature reached by the heater when dissipating 25 W with the surface of the block at a temperature of 35 °C.
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A baseball player throws a ball into the stands at 15.0 m/s and at an angle 45.0° above the horizontal. On its way down, the ball is caught by a spectator 4.10 m above the point where the ball was thrown. How much time did it take for the ball to reach the fan in the stands?

Answers

Answer:

Time = 1.61 seconds

Explanation:

Using the equation displacement of a trajectory motion in the y plane

Y = u t sin ů - ½gt²....equation 1 where

Y= vertical displacement =4.1

U = initial velocity = 15m/s

g = acc. Due to gravity = 10m/s

Ů = angle of trajectory = 45

t = time to reach fan on its way down

Sub into equ 1

4.1 = 15t sinů - ½ * 10t²

4.1 = 10.61t - 5t²

Solve using quadratic formula

t =[-B±( -B² -4AC)^½]/2A....equation 2

Where A = 5, B=10.61, C =4.1

Substitute A,B,C into equ2

t = (10.61±5.53)/10

t = 0.508seconds or 1.61seconds

Since it is on its way down t= 1.61 seconds

A power P is required to do work W in a time interval T. What power is required to do work 3W in a time interval 5T? (a) 3P (b) 5P (c) 3P/5 (a) P (e) 5P/3

Answers

Answer:

(c) 3P/5

Explanation:

The formula to calculate the power is:

P=(W)/(T)

where

W is the work done

T is the time required for the work to be done

In the second part of the problem, we have

Work done: 3W

Time interval: 5T

So the power required is

P=(3W)/(5T)=(3)/(5)(W)/(T)=(3)/(5)P

Air that enters the pleural space during inspiration but is unable to exit during expiration creates a condition called a. open pneumothorax. b. empyema. c. pleural effusion. d. tension pneumothorax.

Answers

Answer:

The correct answer is d. tension pneumothorax.

Explanation:

The increasing build-up of air that is in the pleural space is what we call the tension pneumothorax and this happens due to the lung laceration that lets the air to flee inside the pleural space but it does not return.

As an object in motion becomes heavier, its kinetic energy _____. A. increases exponentially B. decreases exponentially C. increases proportionally D. decreases proportionally

Answers

the answer is c. as an object is in motion speeds up or “becomes heavier”, it’s kinetic energy increases proportionally: double the velocity and you quadruple the kinetic energy. this is why a tiny bullet traveling at high speed does so much more damage than a huge truck bumping into something at 1 mph. so the answer is c

Answer:

c

Explanation:

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

Answers

Answer:

1.8 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:

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.

Sheba is texting and driving. She does not see the baby chinchilasitting in the road! If she was traveling at 35 m/s, and at best her
car can accelerate at -3.5 m/p/s, how far would it take her to stop?

Answers

Answer:

dfmwekognjioq 4eijqgi24gno3p2qtijoq2 gjo23ijgto32tneqew gqewkgnoweqgmnlqwe gwjoegkqwji4toqn34goti34

Explanation:

Other Questions
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