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Two capacitors give an equivalent capacitance of 9.42 pF when connected in parallel and an equivalent capacitance of 1.68 pF when connected in series. What is the capacitance of each capacitor?

Answers

Answer 1

Answer:

Answer:

$C_1=7.23pF\ and\ C_2=2.19pF$

Explanation:

Let the two capacitance are $C_1\ and\ C_2$

It is given that when capacitors are connected in parallel their equvilaent capacitance is 9.42 pF

So $C_1+ C_2=9.2$ --------EQN 1

And when they are connected in series their equivalent capacitance is 1.68 pF

So $(1)/(C_1)+(1)/(C_2)=(1)/(1.68)$

$(C_1+C_2)/(C_1C_2)=(1)/(1.68)$

$C_1C_2=1.68* 9.42=15.8256pF$

$C_1-C_2=√((C_1+C_2)^2-4C_1C_2)=√(9.42^2-4* 15.8256)=5.0432pF$ -----EQN

On solving eqn 1 and eqn 2

$C_1=7.23pF\ and\ C_2=2.19pF$

Related Questions

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A stock person at the local grocery store has a job consisting of the following five segments:1) picking up boxes of tomatoes from the stockroom floor

2)accelerating to a comfortable speed.

3) Carring the boxes to the tomato display at constant speed.

4)decelerating to a stop.

5) lowering the boxes slowly to the floor.

During which of the five segments of the job does the stock person do positive work on the boxes?

A) (2) and (3)

B(1) and (2)

C) (1) only

D) (1), (2), (4) and (5)

E) (1) and (5)

Answers

Answer:

Explanation:

Work done can be said to be positive if the applied force has a component to be in the direction of the displacement and when the angle between the applied force and displacement is positive.

In 1 and 2 work done is positive

an object down, but this is not true. If you place a box of mass 8 kg on a moving horizontal conveyor belt, the friction force of the belt acting on the bottom of the box speeds up the box. At first there is some slipping, until the speed of the box catches up to the speed of the belt, which is 5 m/s. The coefficient of kinetic friction between box and belt is 0.6. (a) How much time does it take for the box to reach this final speed

Answers

Answer:

Explanation:

ASSUMING the belt is horizontal

kinetic friction force is μmg = 0.6(8)(9.8) = 47.04 N

Horizontal acceleration is

a = F/m = 47.04 / 8 = 5.88 m/s²

t = v/a = 5.0 / 5.88 = 0.85034...

t = 0.85 s

A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out. As the man reaches the horizontal section, he grabs a 20.0-kg backpack and skis off a 2.00-m-high ledge. At what horizontal distance from the edge of the ledge does the man land (the man starts at rest)?

Answers

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

$KE_(initial) + PE_(initial) = KE_(final) + PE_(final)$

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

$0 + mgH = (1)/(2)mv^(2) + 0$

$2gH = v^(2)$

$v = √(2* 9.8* 5) = 9.89\ m/s$

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

$mv = (m + m')v'$

$65.0* 9.89 = (65.0 + 20.0)v'$

$v' = 7.56\ m/s$

Now, time taken for the fall:

$h = (1)/(2)gt^(2)$

$t = \sqrt{(2h)/(g)}$

$t = \sqrt{(2* 2)/(9.8) = 0.638\ s$

Now, the horizontal distance is given by:

x = v't = $7.56* 0.638 = 4.823\ m$

Answer

given,

mass of the man = 65 kg

height = 5 m

mass of the back pack = 20 kg

skis off to 2.00 m high ledge

horizontal distance =

speed of the person before they grab back pack is equal to potential and kinetic energy

$mgh= (1)/(2)mv^2$

$v = √(2gh)$

$v = √(2* 9.8 * 5)$

v = 9.89 m/s

now he perform elastic collision

$v = (m_1v_1)/(m_1+m_2)$

$v = (65* 9.89)/(65+20)$

v = 7.57 m/s

time taken by the skies to fall is

$h = (1)/(2)gt^2$

$t = \sqrt{(2h)/(g)}$

$t = \sqrt{(2* 2)/(9.8)}$

t = 0.6388 s

distance

d = v x t

d = 7.57 x 0.6388

d = 4.84 m

Which of the following is a TRUE statement? a. It is possible for heat to flow spontaneously from a hot body to a cold one or from a cold one to a hot one, depending on whether or not the process is reversible or irreversible.
b. It is not possible to convert work entirely into heat.
c. The second law of thermodynamics is a consequence of the first law of thermodynamics.
d. It is impossible to transfer heat from a cooler to a hotter body.
e. All of these statements are false.

Answers

Answer:

e. All of these statements are false.

Explanation:

As we know that heat transfer take place from high temperature to low temperature.

It is possible to convert all work into heat but it is not possible to convert all heat in to work some heat will be reject to the surrounding.

The first law of thermodynamics is the energy conservation law.

Second law of thermodynamics states that it is impossible to construct a device which convert all energy into work without rejecting the heat to the surrounding.

By using heat pump ,heat can transfer from cooler body to the hotter body.

Therefore all the answer is False.

n a distant solar system, a planet of mass 5.0 x 1024 kg orbits a sun of mass 3.0 x 1030 kg at a constant distance of 2.0 x1011 m. How many earth days does it take for the planet ot execute one complete orbit about the sun

Answers

Answer:

F = M2 ω^2 R centripetal force of sun on planet

ω = (F / (M2 R))^1/2 = 2 pi f = 2 pi / P where P is the period

P = 2 pi (M2 * R / F)^1/2

F = G M1 M2 / R^2 gravitational force on planet

P = 2 pi {R^3 / (G M1)]^1/2

P = 6.28 [(2.0E11)^3 / (6.67E-11 * 3.0E30)]^1/2

P = 6.28 (8 / 20)^1/2 E7 = 3.9E7 sec

1 yr = 3600 * 24 * 365 = 3.15E7 sec

P = 3.9 / 3.2 = 1.2 years

A white blood cell has a diameter of approximately 12 micrometers or 0.012 um a model represents its diameter as 24 um what ratio of model size

Answers

Answer:

The ratio of the model size is 1 : 2000

Explanation:

Given

Real Diameter = 0.012 um

Scale Diameter = 24 um

Required

Determine the scale ratio

The scale ratio is calculated as follows;

$Scale = (Real\ Measurement)/(Scale\ Measurement)$

Substitute values for real and scale measurements

$Scale = (0.012\ um)/(24\ um)$

Divide the numerator and the denominator by 0012um

$Scale = (1)/(2000)$

Represent as ratio

$Scale = 1 : 2000$

Hence, the ratio of the model size is 1 : 2000

The ratio of the model size to the actual size is 1 : 2000. This means the model represents the white blood cell's diameter 2000 times larger than its actual size.

The ratio of the model size to the actual size can be calculated using the given measurements:

Actual Diameter = 0.012 um

Model Diameter = 24 um

Ratio = Model Diameter / Actual Diameter

Ratio = 24 um / 0.012 um

Ratio = 2000

So, the ratio of the model size to the actual size is 1 : 2000. This means the model represents the white blood cell's diameter 2000 times larger than its actual size.

Learn more about white blood from the link given below.

brainly.com/question/30712922

#SPJ3

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The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming alpha particles exceeded 32MeV32MeV. This can be explained by the fact that the predictions of the formula apply when the only force involved is the electromagnetic force and will break down if the incoming particles make contact with the nucleus. Use the fact that Rutherford's prediction ceases to be valid for alpha particles with an energy greater than 32MeV32MeV to estimate the radius rrr of the gold nucleus.

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