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pentagon [3]
3 years ago
5

A parallel-plate vacuum capacitor has 8.60 J of energy stored in it. The separation between the plates is 3.80 mm . If the separ

ation is decreased to 1.45 mm , What is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed?
What is the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed?
Physics
1 answer:
creativ13 [48]3 years ago
6 0

Answer:

Part a)

E = \frac{8.60}{2.62} = 3.28 J

Part b)

E = 2.62(8.60) = 22.5 J

Explanation:

As we know that the energy of capacitor when it is not connected to potential source is given as

U = \frac{Q^2}{2C}

As we know that initial energy is given as

8.60 = \frac{Q^2}{2C}

now we know that capacitance of parallel plate capacitor is given as

C = \frac{\epsilon_0A}{d}

now the new capacitance when distance is changed from 3.80 mm to 1.45 mm

C' = \frac{Cd}{d'}

C' = \frac{C(3.80)}{1.45}

C' = 2.62 C

Now the new energy of the capacitor is given as

E = \frac{Q^2}{2(2.62C)}

E = \frac{8.60}{2.62} = 3.28 J

Part b)

Now if the voltage difference between the plates of capacitor is given constant

now the energy energy of capacitor is

U = \frac{1}{2}CV^2

8.60 = \frac{1}{2}CV^2

now when capacitance is changed to new value then new energy is given as

E = \frac{1}{2}C'V^2

E = \frac{1}{2}(2.62C)V^2

E = 2.62(8.60) = 22.5 J

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An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.220 rev/s . The magnitude
matrenka [14]

Answer:

1) The fan's angular velocity after 0.208 seconds is approximately 2.585 rad/s

2) The number of revolutions the blade has travelled in 0.208 s is approximately 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is approximately 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds is approximately 2.312 m/s²

Explanation:

The given parameters are;

The initial velocity of the fan, n = 0.220 rev/s

The magnitude of the angular acceleration = 0.920 rev/s²

The direction of the angular acceleration and the angular velocity = Clockwise

The diameter of the circle formed by the electric ceiling fan blades, D = 0.800 m

1) The initial angular velocity of the fan, ω₀ = 2·π × n = 2·π × 0.220 rev/s = 1.38230076758 rad/s

The angular acceleration of the fan, α = 2·π×0.920 rad/s² = 5.78053048261 rad/s²

The fan's angular velocity, 'ω', after a time t = 0.208 seconds has passed is given as follows;

ω = ω₀ + α·t

From which we have;

ω = 1.38230076758 rad/s + 5.78053048261 rad/s × 0.208 s = 2.58465110796 rad/s

The fan's angular velocity after 0.208 seconds is ω ≈ 2.585 rad/s

2) The number of revolutions the blade has travelled in the given time interval is given from the angle turned, 'θ', in the given time as follows;

θ = ω₀·t + 1/2·α·t²

θ = 1.38230076758 × 0.208 + 1/2 × 5.78053048261 × 0.208² = 0.41256299505 radians

2·π radians = 1 revolution

∴ 0.41256299505 radians = 0.41256299505 radian× 1 revolution/(2·π radian) = 0.06566144 revolution

The number of revolutions the blade has travelled in 0.208 s ≈ 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is given as follows;

The tangential speed, v_t = ω × r = ω × D/2

At t = 0.208 s, ω = 2.58465110796 rad/s, therefore, we have;

v_t = ω × D/2 = 2.58465110796 × 0.800/2 = 1.0338604413

The tangential speed, v_t = 1.0338604413 m/s

The tangential speed ≈ 1.034 m/s

4)  The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds, 'a' is given as follows;

a = α × r = α × D/2

a = 5.78053048261 × 0.800/2 = 2.31221219304

The tangential acceleration, a ≈ 2.312 m/s²

4 0
2 years ago
8. A car initially has 100 J of total energy. After driving down the road, the car's
sveta [45]

The work done on the car is -20 J.

Work done on the car is negative, meaning that the car actually does work on the external system.

<h3>Energy and law of conservation of energy</h3>
  • Energy is the ability to do work
  • the law of conservation of energy states that the total energy in a system is conserved

From the law of conservation of energy, the initial energy of the car before it moves down the road remains constant or unchanged.

  • Initial energy = 100 J
  • Initial energy = Final energy - work done on car
  • Final Energy = Work done on car + initial energy

80J = Work done on car + 100 J

Work done on car = 80 - 100J

Work done on car = -20 J

Hence, the work done on the car is -20 J

Work done on car is negative.

Since work done on the car is negative, it means that the car actually does work on the external system. Hence, the decrease in the energy of the car.

Learn more about energy and work at: brainly.com/question/13387946

8 0
2 years ago
Sam's job at the amusement park is to slow down and bring to a stop the boats in the log ride. If a boat and its riders have a m
shutvik [7]

Answer:

Sam will do 1152 J of work to stop the boat

Explanation:

Work: This is defined as the product of force and distance, the S.I unit of work is Joules. At any point in science, during calculation Energy and worked can be interchange because they have the same unit.

E = W = 1/2mv²................ Equation 1

Where E = energy, W = work, m = mass, v = velocity.

Given: m = 900 kg, v = 1.6 m/s

Substituting these values into equation 1

W = 1/2(900)(1.6)²

W = 450×2.56

W = 1152 J.

Therefore Sam will do 1152 J of work to stop the boat

7 0
3 years ago
A truck moving at 13.3 m/s hits a concrete wall. As a result of the collision, a 6-kg wrench moves forwards and strikes the wall
Naddik [55]

Answer:

Explanation:

The velocity of the wrench must be equal to the velocity of the truck . So momentum of the wrench before it hits the wall

= mv = 6 x 13.3 = 79.8 kg m /s

If resisting force of wall be F , impulse on the wrench = F x time

= F x .07

Impulse = change in momentum of the wrench = mv - 0 = mv = 79.8 kgm/s

So F x .07 = 79.8

F = 1140 N .

8 0
3 years ago
Does a spring scale measure weight or mass? Why?
Marat540 [252]
A spring scale measures weight because <span>It works by Hooke's Law, which states that the force needed to extend a </span>spring<span> is proportional to the distance that </span>spring<span> is extended from its rest position. Therefore, the </span>scale<span> markings on the </span>spring<span> balance are equally spaced. A </span>spring scale<span> can</span>not measure mass<span>, only </span>weight<span>. hope that helped</span>
5 0
3 years ago
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