Number of conducting plates of a multiplate capacitor is 5. The no. Of capacitors is
A.1
B.2
C.3
D.4
(Ans with explanation pls)

Answers

Answer 1

Answer:

4 capacitors

Explanation:

Given

[tex]n = 5[/tex] --- conducting plates

Required

The number of capacitor (c)

This is calculated as:

[tex]c = n - 1[/tex]

So, we have:

[tex]c = 5 - 1[/tex]

[tex]c = 4[/tex]


Related Questions

En el proceso de diseño de ingeniería, ¿qué limitaciones deben tenerse en cuenta al utilizar un modelo / prototipo?

Answers

Answer:

Las limitaciones de un modelo o prototipo son;

1) Los parámetros ambientales (donde se opera el modelo, prototipo o producto) son diferentes y, por lo tanto, pueden producir relaciones y factores ambientales que serán diferentes de los factores ambientales y las relaciones del objeto real.

2) El análisis del problema puede ser inadecuado

3) La posibilidad de falta de satisfacción del cliente con un modelo, preferencia por la demostración real del producto.

4) Reproducción inexacta del entorno del producto durante la prueba del modelo

5) El factor de costo del modelo

6) Mayor complejidad introducida por el modelo / prototipo al análisis de la solución

Explanation:

El modelo o prototipo es la presentación del diseño articulado, construido para demostrar el producto real con el propósito de encontrar la existencia de errores en el diseño que serían corregidos, antes de que se realice la producción real

DEFINE UNIFORM AND NON UNIFORM VELOCITY

Answers

Explanation:

Uniform velocity is when an object goes an equal amount of space in an equal amount of time whereas non uniform velocity is when the object covers an unequal amount of distance in an equal amount of time.

1o4 Fahrenheit is equal to​

Answers

Answer:

40 degrees Celsius

Explanation:

Have a great summer :)

40 degrees Celsius is the answer your looking for : )

What is energy and what is the formula of energy

Answers

Answer:

Energy is: the ability to do work

The formula for energy is: power x time = energy

Explanation:

Hope this helps (there isn't really an explanation)

calculate:An elcetric lamp is rated 240V and 40W. What is the cost of running the lamp for 62 hours if the elctricity authorities charges #2.50k per KWH

Answers

Explanation:

The voltage of the lamp, V = 240 V

Power of the lamp, P = 40 W

It is running for 62 hours.

The cost of running is $2.50k per KWH

Electric power is,

P = 40×62 Wh

= 2480 Wh

P = 2.48 kWh

At the rate of $2.5 per kWh

P = $6.2

So, the cost of running is $6.2 per kWh.

When the interval between the stimuli decreases, _______.
A. a second action potential is generated until the interval reaches the absolute refractory period
B. a second action potential is generated regardless of the stimulus and the interval
C. a second action potential is generated until the interval reaches the relative refractory period
D. a second action potential is generated as long as the stimulus is above threshold

Answers

Answer:

The correct option is A. a second action potential is generated until the interval reaches the absolute refractory period.

Explanation:

The inter-stimulus interval (ISI) is the temporal interval between two successive stimuli, measured from the offset of the first stimulus to the commencement of the second.

A cell's refractory period is the time during which it is unable to replicate an action potential. Therefore, the absolute refractory period is the amount of time it takes for a second action potential to be initiated, regardless of how large a stimulus is applied repeatedly.

A second action potential is generated when the gap between the stimuli decreases until the interval reaches the absolute refractory period.

Therefore, the correct option is A. a second action potential is generated until the interval reaches the absolute refractory period.

That is, when the interval between the stimuli decreases, a second action potential is generated until the interval reaches the absolute refractory period.


Two friends are working out together using weights:
-James lifts 20kg weights up at a velocity of 1.9m/s
-Kayla lifts 15kg weights up at a velocity of 2.6m/s
0
Which friend is creating more momentum?

Answers

Answer:

Kayla. You can calculate it using the formula for momentum: momentum=mass×velocity and find the bigger number between the two momentums

.If a vehicle covers 3 km in 5 minutes, calculate the speed of the vehicle? (With process )

Answers

Answer:

Speed = 1.6 m/s

Explanation:

Formula,

Speed = Distance ÷ Time

I need this now

On the planet Xenos, an astronaut observes that a 2.0 m long pendulum has a period of 2.2 s.
What is the free-fall acceleration on Xenos? *

Answers

Answer:

[tex]g=16.31\ m/s^2[/tex]

Explanation:

Given that,

The length of the pendulum, l = 2 m

The period of the pendulum, T = 2.2 s

The formula for the time period of a pendulum is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

or

[tex]T^2=4\pi ^2\dfrac{l}{g}\\\\g=\dfrac{4\pi ^2l}{T^2}\\\\g=\dfrac{4\pi ^2\times 2}{(2.2)^2}\\\\g=16.31\ m/s^2[/tex]

So, the free fall acceleration is [tex]16.31\ m/s^2[/tex].

How does an electric bulb work?
this is a class 6 question...​

Answers

Answer:

Explanation:

When a light bulb connects to an electrical power supply, an electrical current flows from one metal contact to the other. As the current travels through the wires and the filament, the filament heats up to the point where it begins to emit photons, which are small packets of visible light.

A 500-kg crate sits on a 10-degree ramp. If friction between the ramp and the crate is 800 N, what is the acceleration of the crate?

Answers

By Newton's second law, the net force acting on the crate parallel to the surface is

F = mg sin(10°) - 800 N = ma

where m = 500 kg is the mass of the crate and a is the acceleration.

Solve for a :

a = ((500 kg) (9.80 m/s^2) sin(10°) - 800 N) / (500 kg)

a ≈ 0.102 m/s^2

what two forces contribute to the nuclear tug of war in an atom

Answers

Answer:2 protons and 2 neutrons

Explanation:In Nuclei, There are 2 forces. 1 force is electrostatic and acts as repulsion between 2 protons. The other is force of attraction called Nuclear force between 2 neutrons.

A fixed mass of gas has a volume of gas of 25cm3. the pressure of the gas is 100kPA. the volume of the gas is slowly decreased by 15cm3 at a constant temperature. what is the change in the pressure of the gas?
a) 67kPA
b) 150kPA
c) 170kPA
d) 250kPA
give reasons

Answers

A fixed mass of gas has a volume of 25 [tex]cm^3[/tex], the pressure of the gas is 100 kPa, the volume of the gas is slowly decreased by 15 [tex]cm^3[/tex] at a constant temperature, and the change in pressure of the gas is 150 kPa, which is option b.

What is the calculation of the change in pressure?

PV = nRT (P= pressure of the gas, V =volume, n = number of moles of gas, R = gas constant, and T =temperature of the gas in kelvin)

Suppose the gas is an ideal gas and that the temperature is constant,

P1V1 = P2V2

Here P1 = 100 kPa, V1 = 25 [tex]cm^3[/tex], V2 = 10 [tex]cm^3[/tex],

100 kPa x 25 [tex]cm^3[/tex] = P2 x 10 [tex]cm^3[/tex]

P2 = (100 kPa x 25 [tex]cm^3[/tex]) / 10 [tex]cm^3[/tex]

P2 = 250 kPa

the change in pressure of the gas is,

ΔP = P2 - P1 = 250 kPa - 100 kPa = 150 kPa

The reason is that when the volume of a fixed mass of gas is decreased, the pressure of the gas increases proportionally, so here assuming that the temperature is constant it is calculated.

Hence, the volume of the gas is slowly decreased by 15 [tex]cm^3[/tex] at a constant temperature, and the change in pressure of the gas is 150 kPa, which is option b.

Learn more about the calculation of the change in pressure here.

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A woman pushes a car with a force of 400 N for a distance of 15m. How much work has she done?

Answers

Answer:

the work done by the woman is 6,000 J

Explanation:

Given;

force applied by the woman, F = 400 N

distance moved by the woman, d = 15 m

The work done by the woman is calculated as follows;

W = F x d

W = 400 N x 15 m

W = 6000 Nm = 6,000 J

Therefore, the work done by the woman is 6,000 J

What is the gravitational force between two objects?​

Answers

Answer:

Gravitational force -an attractive force that exists between all objects with mass; an object with mass attracts another object with mass; the magnitude of the force is directly proportional to the masses of the two objects and inversely proportional to the square of the distance between the two objects.

Explanation:

hope it helped

Gravitational force

Temperature of substance changes from -20 celsius to 20 celsius. What is temperature change in kelvin scale

Answers

Answer:

313kelvin

Explanation:

40 degree celcius plus 273=313K

Find the acceleration a body whose velocity increases from 11m/s to 33m/s in 10 seconds

Answers

Answer:

I am not sure if this is the answer

acceleration: 2.2m/s

Explanation:

here

initial velocity(u): 11m/s

Final velocity(v): 33m/s

time taken(t): 10 s

now

a:v-u/t

or

acceleration:final velocity-initial velocity/time taken

or

a: 33-11/10

or

a:22/10, divide it

: a=2.2m/s#

A force of 20000N acts on the raft in the direction down
State the name given to the force shown by arrow in Fig.
Calculate the mass of the raft.

Answers

Answer:

Figure is not there

Explanation:

what is gravity..
what is force.
mention the two type of force
and give 3,3 exmples

Answers

Answer:

Oh, umm…

Jump! (The higher, the better!)

Drop your pencil! (or pen or ruler, whichever you prefer!)

Throw a ball of paper! (and make sure you pick it back up later!)

Explanation:

if you satisfied to my answer , just put brainliest please , and ur welcomeee♥️♥️

You throw a water balloon straight up with a velocity of 13 m/s. What is its
maximum height?
O A. 4.4 m
B. 6.3 m
C. 10.7 m
D. 8.6 m

Answers

Answer:

Explanation:

[tex]h=-v^2 /2g[/tex]

[tex]with\\g = 9,8 m/s^2 or 10 m/s^2[/tex]

[tex]h= (-13)^2 / 2 * 9,8 = 8,6[/tex]

When you flip a penny (2.35 g), it leaves your hand and moves upward at 2.85 m/s. Use energy to find how high the penny goes above your hand before stopping. A (b) The penny then falls to the floor, 1.26 m below your hand. Use energy to find its speed just before it hits the floor. A (c) Explain your choice of reference level for parts (a) and (b). C (d) Choose a different reference level and repeat part (b)

Answers

Answer:

a. 0.41 m

b. 5.72 m/s

c. i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.

ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.

d. 5.72 m/s

Explanation:

a. Use energy to find how high the penny goes above your hand before stopping.

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy at the hand, E equals the total mechanical energy when the penny stops in the air, E'.

E = E'

U + K = U' + K' where U = initial potential energy at hand level = mgh where h = height at hand level = 0, K = initial kinetic energy at hand level = 1/2mv² where v = speed at hand level = 2.85 m/s, U' = final potential energy at stopping level = mgh' where h' = height at stopping level, K = final kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops)

So, U + K = U' + K'

mgh + 1/2mv² = mgh' + 1/2mv'²

substituting the values of the variables into the equation, we have

mg(0) + 1/2m(2.85 m/s)² = mgh' + 1/2m(0 m/s)²

0 + 1/2m(8.1225 m²/s²) = mgh' + 0

m(4.06125 m²/s²) = mgh'

h' = 4.06125 m²/s² ÷ g

h' = 4.06125 m²/s² ÷ 9.8 m/s²

h' = 0.41 m

(b) The penny then falls to the floor, 1.26 m below your hand. Use energy to find its speed just before it hits the floor.  

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E'  equals the total mechanical energy on the ground, E"

E' = E"

U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh" where h' = height at stopping level = height of penny above hand, h' + height of hand above ground = 0.41 m + 1.26 m = 1.67 m, K = initial kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₁ = height at ground level = 0, K = final kinetic energy at ground level = 1/2mv"² where v" = speed at ground level,

So, U' + K' = U' + K'

mgh" + 1/2mv'² = mgh₁ + 1/2mv"²

substituting the values of the variables into the equation, we have

mg(1.67 m) + 1/2m(0 m/s)² = mg(0) + 1/2mv"²

1.67mg + 0 = 0 + 1/2mv"²

1.67mg = 1/2mv"²

1.67g = 1/2v"²

v"² = 2(1.67g)

v" = √[2(1.67g)]

v" = √[2(1.67 m × 9.8 m/s²)]

v" = √[2(16.366 m²/s²)]

v" = √[32.732 m²/s²)]

v" = 5.72 m/s

(c) Explain your choice of reference level for parts (a) and (b).

i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.

ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.

(d) Choose a different reference level and repeat part (b)

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E'  equals the total mechanical energy on the ground, E"

E' = E"

U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh' where h' = height at stopping level = 0.41 m, K = initial kinetic energy at stopping level = 1/2mv'² where v' = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₂ = height of hand above the ground level = height of ground below hand = -1.26 m(it is negative since the ground is below the hand), K = final kinetic energy at ground level = 1/2mv"² where v = speed at ground level,

So, U' + K' = U' + K'

mgh' + 1/2mv'² = mgh₂ + 1/2mv"²

substituting the values of the variables into the equation, we have

mg(0.41 m) + 1/2m(0 m/s)² = mg(-1.26 m) + 1/2mv"²

0.41mg + 0 = -1.26 mg + 1/2mv"²

0.41mg + 1.26mg = 1/2mv"²

1.67mg = 1/2mv"²

1.67g = 1/2v"²

v"² = 2(1.67g)

v" = √[2(1.67g)]

v" = √[2(1.67 m × 9.8 m/s²)]

v" = √[2(16.366 m²/s²)]

v" = √[32.732 m²/s²)]

v" = 5.72 m/s

what is fundamental​

Answers

Fundamental

forming a necessary base or core; of central importance.

"the protection of fundamental human rights"

If the car falls down the side of the cliff, what is happening to the gravitational potential energy of the falling car (Assume the bottom of the cliff is zero)
Group of answer choices

the gravitational potential energy is decreasing

the gravitational potential energy has not changed

the gravitational potential energy is increasing

Answers

Explanation:

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. ... The gravitational potential energy is equal to its weight times the height to which it is lifted. PE = kg x 9.8 m/s2 x m = joules. The 9.8 us the gravitational acceleration constant.

so the answer is "the gravitational potential energy is decreasing"

6) A car of mass 1000kg moving with a velocity of 40m/s collides with a tree and comes to stop in 5s. What will be the force exerted by the car on the tree?

Answers

Answer:

Refer to the attachment!~


Which two forms of electromagnetic energy are used to produce the
most spectacular fluorescence when placed in darkness?
1. microwaves and x rays
2. microwaves and infrared
3. ultraviolet and x rays
4. ultraviolet and infrared

Answers

Answer:

3. ultraviolet and x rays

The two forms of electromagnetic energy are used to produce the most spectacular fluorescence when placed in darkness are ultraviolet and x rays.

What is electromagnetic energy?

Electromagnetic radiation is waves of the electromagnetic field, propagating through space, carrying electromagnetic radiant energy. It includes radio waves, microwaves, infrared, light, ultraviolet, X-rays, and gamma rays.

What is fluorescence?

Fluorescence is the process in which a substance absorbs light at a high energy, short wavelength and emits light at a lower energy, usually visible wavelength.

What is ultraviolet rays?

Ultraviolet rays is a type of electromagnetic waves in which the wavelength is shorter than visible rays. It is responsible for 10% of sunlight and causes sun tan. It is used to purify water in water purifiers. It kill germs.

What is x rays?

X rays lies beyond ultraviolet rays. It is used to diagnose in medical field. It can destroy living tissues so excessive exposure should be limited to reduce harmful effect.

The color change of fluorescent minerals is most spectacular when the minerals are placed in darkness and exposed to electromagnetic energy shorter than visible light.

Ultraviolet and x rays are shorter than visible light so they exhibit fluorescent property.

To learn more about Electromagnetic radiation here

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A.All three bulbs will go out
B.Bulbs 2 and 3 will go out, but bulb 1 will remain lit
C. All three bulbs will go out
D. Bulb 3 will go out, but bulbs 1 and 2 remain lit

Answers

Answer:

(D)

Explanation:

When switch C is opened then, Current is not flowing across 3 So bulb 3 will go out.

But current is flowing across 1 and 2 bulb because their switch is closed

therefore bulb 1 and 2 will remain it.

Hence, option (D) will be correct.

explain how conservation of mechanical energy occurs a kicking a ball

Answers

The law of conservation of energy says that energy cannot be created or destroyed. This means that even though energy changes form, the total amount of energy always stays the same. Your muscles move your leg, your foot kicks the ball, and the ball gains kinetic energy from the kick

10.
You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is
negligible friction between your feet and the ice. A friend throws you a 0.4 kg ball that is traveling
horizontally at 10 m/s. Your mass is 70 kg. If you catch the ball, with what speed do you and the ball
move afterwards?

1.02 m/s

0.06 m/s

0.02 m/s

0.12 m/s

Answers

Answer:

Explanation:

This is a classic Law of Momentum Conservation problem. For us the equation will look like this:

[tex][(m_yv_y+m_bv_b)]_b=[(m_y+m_b)v_{both}]_a[/tex] Filling in with our given info:

[tex][(70.0)(0)+(.40)(10.0)]_b=[(70.0+.40)v_{both}]_a[/tex] and

4.0 = 70.4v and

v = .06 m/s

Complete the equation to show the radioactive decay of carbon-14 to nitrogen-14

Answers

Answer:

The beta decay takes place.

Explanation:

The reaction of radioactivity of carbon 14 to nitrogen 14 is

There is a beta decay.  

The reaction is

[tex]C_{6}^{14}\rightarrow N_{7}^{14}+\beta _{-1}^{0}+ energy[/tex]

Here some energy is released in form of neutrino.

A road with a radius of 75.0 m is banked so that a car can navigate the curve at a speed of 15.0 m/s without any friction. When a car is going 31.8 m/s on this curve, what minimum coefficient of static friction is needed if the car is to navigate the curve without slipping?

Answers

Find the angle θ made by the road. When rounding the curve at 15.0 m/s, the car has a radial acceleration of

a = (15.0 m/s)² / (75.0 m) = 3.00 m/s²

There are two forces acting on the car in this situation:

• the normal force of the road pushing upward on the car, perpendicular to the surface of the road, with magnitude n

• the car's weight, pointing directly downward; its magnitude is mg (where m is the mass of the car and g is the acceleration due to gravity), and hence its perpendicular and parallel components are, respectively, -mg cos(θ) and mg sin(θ)

By Newton's second law, the net forces in the perpendicular and parallel directions are

(perp.) ∑ F = n - mg cos(θ) = 0

(para.) ∑ F = mg sin(θ) = ma

==>   sin(θ) = a/g   ==>   θ = arcsin(a/g) ≈ 17.8°

(Notice that in the paralell case, the positive direction points toward the center of the curve.)

When rounding the curve at 31.8 m/s, the car's radial acceleration changes to

a = (31.8 m/s)² / (75.0 m) ≈ 13.5 m/s²

and there is now static friction (mag. f = µn, where µ is the coefficient of static friction) acting on the car and keeping from sliding off the road, hence pointing toward the center of the curve and acting in the parallel direction. Newton's second law gives the same equations, with an additional term in the parallel case:

(perp.) ∑ F = n - mg cos(θ) = 0

(para.) ∑ F = mg sin(θ) + f = ma

The first equation gives

n = mg cos(θ)

and substituting into the second equation, we get

mg sin(θ) + µmg cos(θ) = ma

==>   µ = (a - g sin(θ)) / (g cos(θ)) = a/g sec(θ) - tan(θ) ≈ 1.12

Answer:

Explanation:

You are in the chapter on Physics about uniform circular motion and gravity. This is a centripetal force problem in particular, and the equation for that is

[tex]F_c=\frac{mv^2}{r}[/tex] where

[tex]F_c[/tex] is the centripetal force needed to keep the car moving in its circular path,

m is the mass of the car,

v is the velocity with which the car is moving, and

r is the radius of the circle that the car is moving around.

For us, the centripetal force is supplied by the friction keeping the car on the road, altering the equation to become

[tex]f=\frac{mv^2}{r}[/tex] and friction is defined by

f = μ[tex]F_n[/tex] (the coefficient of friction multiplied by the weight of the car).

Going on and getting buried even deeper,

[tex]F_n=mg[/tex] which says that the weight of the car is equal to its mass times the pull of gravity. Putting all that together, finally, we have the equation we need to solve this problem:

μ·m·g = [tex]\frac{mv^2}{r}[/tex] and we solve this for μ:

μ = [tex]\frac{mv^2}{mgr}[/tex] and it just so happens that the mass of the car cancels out. (I'll tell you why the mass of the car doesn't matter at the end of this problem). Filling in and solving for the coefficient of friction:

μ = [tex]\frac{31.8^2}{(9.8)(75.0)}[/tex] to 2 significant figures is

μ = 1.4

The mass of the car doesn't affect whether or not the car can stay on the curve. Even though a car with a greater mass will have a greater frictional force, that doesn't mean that it's easier for that car to stay on the road; a larger mass only means that a larger centripetal force is needed to keep it moving in a circle. This makes the gain in friction become offset by the fact that a larger centripetal force is necessary. Thus,

On a flat curve, the mass of the object experiencing circular motion does not affect the velocity at which it can stay on the curve.

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