Power=170 watts time=20 seconds

C1=4F, C2=4F, C3=2F, C4=4F, C5= 9.2 F. Calculate the equivalent capacitance

grin007 [14]

The equivalent capacitance between A and B points is 2.5F.

<h3>What is parallel plate capacitor?</h3>

The two parallel plates placed at a distance apart used to store charge when electric supply is on.

The capacitance of a capacitor is given by

C = ε₀ A/d

From the given circuit C1, C2 and C3, C4 are in parallel C1=4F, C2=4F, C3=2F, C4=4F, C5= 9.2 F

C1, C2 = 4 +4 =8F

C3, C4 = 2 +4 =6F

Now , all capacitors are in series.

Total equivalent capacitance is
1 / Ceq = 1/ 8 +1/6 +1/ 9.2

Ceq = 2.5 F

Thus, the equivalent capacitance between A and B points is 2.5F.

Learn more about parallel plate capacitor.

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6 0

2 years ago

A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav

gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

$K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2$ (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

$v^2=v_o^2+2gy$

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}$

With this value you can find the spring constant k from the equation (1):

$mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}$

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

$W=F_e\\\\mg=kx$

you solve the last expression for x:

$x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m$

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}$

Next, you calculate x from the equation (1):

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m$

The net fire is stretched 2.72 m

5 0

3 years ago

Two technicians are discussing electromagnetic induction. Technician A says that the induced voltage can be increased if the spe

Inga [223]

Answer:Both

Explanation:

There are three ways to increase the induced voltage in electromagnetic induction:

1) increase the speed at which the conductor moves through the magnetic field. This means that the lines of flux are cut more quickly and more emf is induced.

2) use stronger magnets which provides a stronger magnetic field and more densely packed lines of flux.

3) use a coil of multiple loops.

Hence both technicians were correct.

3 0

3 years ago

A battery charger can produce 3A at 12 Volt and charges a battery fer 2 hr. Calculate work in KJ.

Oksi-84 [34.3K]

Answer: 259.2 KJ

Explanation:

The formula calculate work don in a circuit is given by :-

$W=QV$ , where Q is charge and V is the potential difference.

The formula to calculate charge in circuit :-

$Q=It$ , where I is current and t is time.

Given : Current : $I=3A$

Potential difference : $V=12\ V$

Time : $t=2\ hr=2(3600)\text{ seconds}=7200\text{ seconds}$

Now, $Q=3(7200)=21,600\ C$

Then, $W=(21600)(12)=259,200\text{ Joules}=259.2\text{ KJ}$

Hence, the work done = 259.2 KJ

4 0

3 years ago

A ramp leading to the entrance of a building is inclied upward at an angle of 7 degree. A suitcase is to be pulled up the ramp b

photoshop1234 [79]

Explanation:

The angle of the handle relative to the horizontal is 35°. The angle of the ramp to the horizontal is 7°. So the angle of the handle relative to the ramp is 28°.

cos 28° = 50 / F

F = 50 / cos 28°

F = 56.6 lbs

8 0

3 years ago