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Leona [35]
4 years ago
5

An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and ind

uctor, or a capacitor. At time t = 0 the voltage is zero. At time t = T/4 the current in the unknown element is equal to zero, and at time t = T/2 the current is I = -Imax, where Imax is the current amplitude. What is the unknown element?
Physics
1 answer:
torisob [31]4 years ago
6 0

Answer:

This satisfy the above given condition so we can say that this capacitor.

Explanation:

Let's take one by one option and check whether is wrong or right

<u>For inductor:</u>

I=I_osin(wt-\frac{\pi }{2})

Given that at t=T/4 ,I=0 and we know that

w=\dfrac{2\pi }{T}

So at T/4

I=I_osin(\frac{2\pi }{T}\times \frac{T}{4}-\frac{\pi }{2})

I=0 A

At t=T/2

I=I_osin(\frac{2\pi }{T}\times \frac{T}{2}-\frac{\pi }{2})

I=I_o

It means that this not a indutor.

<u>For capacitor:</u>

I=I_osin(wt+\frac{\pi }{2})

At T/4, I=0

At t=T/2

I=I_osin(\frac{2\pi }{T}\times \frac{T}{2}+\frac{\pi }{2})

I= -I_o

This satisfy the above given condition so we can say that this capacitor.

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Cherise sets identical magnetic carts on two tracks. At the end of each track is a blue magnet that cannot move. Cherise can mov
Colt1911 [192]

Answer:

it's d

Explanation:

7 0
3 years ago
A crate of oranges is shoved across a grocery store floor with a force of 100 N for a distance of 1 meter. The crate travels an
navik [9.2K]

The statements that are true are;

  • the magnitude of work done by frictional forces on the crate is 100 J
  • the magnitude of work done by the applied force on the crate is 100 J
<h3>What is work done?</h3>

Work is said to be done when te force applied travels a distance in the direction of the force. Now we can see that when the force is applied by shoving the crate, work is done, an additional work is done by friction to bring the crate to a stop.

Hence, the following are true;

  • the magnitude of work done by frictional forces on the crate is 100 J
  • the magnitude of work done by the applied force on the crate is 100 J

Learn more about work: brainly.com/question/18094932?

4 0
2 years ago
A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine s
Ierofanga [76]

Answer:

a). P=11.04kW

b). Pmax=11.38 kW

c). Wt=6423.166kJ

Explanation:

The power of the motor when the speed is constant is the work in a determinate time.

P=\frac{W}{t}

The work is the force the is applicated in a distance so

W=F*d

replacing:

P=F*\frac{d}{t} and \frac{d}{t} determinate distance in time is velocity so

a).

P=F*v

F=m*a\\F=m*g*sin(33.5)

P=950kg*9.8\frac{m}{s^{2}}*sin(33.5)*2.15\frac{m}{s}\\P=11047.846 W\\P=11.0478 kW

b).

The maximum power must the motor provide, is the maximum force with the maximum speed of the motor in this case

The first step is find the acceleration so

vi=0\frac{m}{s} \\vf=2.15 \frac{m}{s}\\vf=vi+a*t\\vf-vi=a*t\\ a=\frac{vf-vi}{t}= a=\frac{2.15\frac{m}{s}-0\frac{m}{s}}{13s}\\a=0.1653 \frac{m}{s^{2}}

The maximum force is when the car is accelerating so

Ft=Fa+Fg\\Ft=m*a+m*g*sin(33.5)\\Ft=950kg*0.1653\frac{m}{s^{2}}+950*9.8\frac{m}{s^{2}}*sin(33.5)\\Ft=5295.565 N

so the maximum force is the maximum force by the maximum speed

Pmax=Ft*v\\Pmax=5295.565N*2.15\frac{m}{s}\\Pmax=11385.46\\Pmax=11.3854kW

c).

The total energy transfer without any friction is the weight move in the high axis y in this case, so is easy to know that distance

W=m*g*h

h=Length*sin(33.5)

W=m*g*Length*sin(33.5)

W=950 kg*9.8* 1250m*sin(33.5)

W=6423166.667 kJ

W=6423.166 kJ

4 0
4 years ago
1. 3 main components of a circuit
siniylev [52]
Voltage, resistance and current are the three components that must be present for a circuit to exist. A circuit will not be able to function without these three components. Voltage is the main electrical source that is present in a circuit. :)
7 0
3 years ago
Read 2 more answers
An electric field of 8.20 ✕ 105 V/m is desired between two parallel plates, each of area 25.0 cm2 and separated by 2.45 mm. Ther
astraxan [27]

Answer:

q = 1.815 \times 10^{-8} C

Charge on one plate is positive in nature and on the other plate it is negative in nature.

Explanation:

E = 8.20 x 10^5 V/m, A = 25 cm^2, d = 22.45 mm

According to the Gauss's theorem in electrostatics

The electric field between the two plates

E = \frac{\sigma }{\varepsilon _{0}}

{\sigma }= E \times {\varepsilon _{0}}

{\sigma }= 8.20 \times 10^{5} \times {8.854 \times 10^{-12}

{\sigma }= 7.26 \times 10^{-6} C/m^{2}

Charge, q = surface charge density x area

q = 7.26 \times 10^{-6} \times 25 \times 10^{-4}

q = 1.815 \times 10^{-8} C

5 0
3 years ago
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