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Drupady [299]
3 years ago
9

A 70 pF capacitor and a 280 pF capacitor are both charged to 3.6 kV. They are then disconnected from the voltage source and are

connected together, positive plate to positive plate and negative plate to negative plate.
Find the resulting potential difference across each capacitor.
kV (70 pF capacitor)
kV (280 pF capacitor)
Find the energy lost when the connections are made.J
Physics
1 answer:
e-lub [12.9K]3 years ago
6 0

Answer:

0.636 kJ

Explanation:

The charge on any capacity, q = CV, thus,

The initial charge on the 70 pF capacitor is

q = Cv

q = 70*10^-12 * 3.6*10^3

q = 2.52*10^-7 C

The charge on the 280 pF capacitor is q = C*v

q = 280*10^-12 * 3.6*10^3

q = 1.008x10^-6 C

When they are connected as stated, the net total charge remaining will be 1.008*10^-6 - 2.52*10^-7 = 7.56*10^-7 C

Since the capacitors are in parallel, the equivalent capacitance will be 70 + 280 pF = 350 pF

Remember, q = CV, then V = q/C

V = 7.56*10^-7 C / 350*10^12 F

V = 2160 V

b) The energy before is 1/2 C*v²

E = 1/2 * 70*10^-12 * 3600² + 1/2 * 280*10^-12 * 3600²

E = 4.536*10^-4 J + 1.814*10^-3 J

E = 2.268 kJ

The energy After is 1/2 Cv²

E = 1/2 * 70*10^-12 * 2160² + 1/2 * 280*10^-12 * 2160²

E = 3.266*10^-4 J + 1.306*10^-3 J

E = 1.632 kJ

so the loss is 2.268 - 1.632 = 0.636 kJ

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bazaltina [42]

Answer:

Explanation:

The movement of a body can be analyzed using New's first law. In an inertial frame (without acceleration) every body is kept at rest or moving at constant speed until there is an external force that changes this state

Let's analyze these cases in the framework of this first law

a) If the vehicle is going at constant speed the two bodies (the egg and the hands) do not change movement so he had returned to the hands

b) If the vehicle accelerates the passenger goes faster, but the egg that is not subject to anything does not change the movement, so it falls behind the passenger

c) If the vehicle slows down, the passenger reduces its speed and the distance traveled in time, but the egg that is not attached follows its movement and falls in front of the passenger.

4 0
3 years ago
A 100-lb child stands on a scale while riding in an elevator. What does the scale read while the elevator slows to stop at the l
Lelechka [254]

Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor

Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.

<h3>What is the apparent weight of a body in a lift?</h3>
  • Consider a body of mass m kept on a weighing machine in a lift.
  • The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
  • The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
<h3>How to solve the question?</h3>
  • Here we have given with the actual weight of the body as 100lbs.
  • This 100lb child is standing on the scale or the weighing machine, when it is riding .
  • During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
  • There is also<em> mg </em>downwards and a normal reaction in the upward direction.
  • when we equate both the upward force and downward force, we get,

                             ma=mg-N\\N=mg-ma    i.e. during riding the scale reads a weight less than that of actual weight.

  • When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.

Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.

Learn more about the apparent weight of the body in a lift here:

brainly.com/question/28045397

#SPJ4

7 0
1 year ago
A(n) 2602 kg van runs into the back of a(n)
forsale [732]

Answer:

8.5 m/s

Explanation:

please see paper for the work!

8 0
2 years ago
Water enters a student's house 10.0 m above the ground through a pipe with a cross section area of 1.00 x 10-4m2 at ground. Insi
dezoksy [38]

Answer:

(a). V₁ = 10m/s (velocity inside the house), V₂ = 5m/s (velocity at ground level)

(b). P₂ = 236500 Pa

Explanation:

This is quite straight-forward so let us begin by defining the terms given.

Given that;

The cross-section area inside the student's house A₁ = 0.50 0.50 x 10-4m2.

Let us make the velocity of water inside the house be V₁

such that the Volume of water entering the per second is = A₁V₁

Therefore, in 90sec:

45 L =  90 A₁V₁

V₁ = 45 * 10⁻³m³ / 90*0.5*10⁻⁴

V₁ = 10m/s            (velocity of water inside the house)

From the continuity equation we have that;

A₁V₁ = A₂V₂

0.5*10⁻⁴ * 10 = 1*10⁻⁴ V₂

V₂ = 5m/s               (velocity at ground level)

(b). We are told to calculate the water pressure in the pipeline at the ground level.

Using Bernoulli's equation;

P₁ + pgh₁ + 1/2PV₁²  (inside)      =       P₂ + pgh₂ + 1/2PV₂²   (ground level)

1.01*10⁵ + 1000*9.8*10 + 1/2*1000*(10)² = P₂ + 0 + 1/2*1000*(5)²

P₂ (pressure) = 1.01*10⁵Pa

Therefore we have;

101000 + 98000 + 50000 = P₂ + 12500

P₂ = 236500 Pa

cheers I hope this helped !!

3 0
3 years ago
ITS TIMED PLEASE HELP !!!!!
dusya [7]

Answer: direct linear

Explanation:

3 0
3 years ago
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