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Marina CMI [18]
3 years ago
10

The capacitor is now disconnected from the battery, and the plates of the capacitor are then slowly pulled apart until the separ

ation reaches 3d3d. Find the new energy U1U1U_1 of the capacitor after this process.
Physics
1 answer:
alexdok [17]3 years ago
3 0

Answer:

U₁ = (ϵAV²)/6d

This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.

Explanation:

The energy stored in a capacitor is given by (1/2) (CV²)

Energy in the capacitor initially

U = CV²/2

V = voltage across the plates of the capacitor

C = capacitance of the capacitor

But the capacitance of a capacitor depends on the geometry of the capacitor is given by

C = ϵA/d

ϵ = Absolute permissivity of the dielectric material

A = Cross sectional Area of the capacitor

d = separation between the capacitor

So,

U = CV²/2

Substituting for C

U = ϵAV²/2d

Now, for U₁, the new distance between plates, d₁ = 3d

U₁ = ϵAV²/2d₁

U₁ = ϵAV²/(2(3d))

U₁ = (ϵAV²)/6d

This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.

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3 0
3 years ago
Read 2 more answers
Solve the equation<br> (2.40 x 10^7)* (3.10 x 10-5^-5)
Mariulka [41]

Explanation:

(2.40 × 10⁷) × (3.10 × 10⁻⁵)

(2.40 × 3.10) × (10⁷ × 10⁻⁵)

7.44 × 10²

7 0
4 years ago
An air track car with a mass of 0.75 kg and a velocity of 8.5 m/s to the right collides elastically with a 0.65kg car moving to
Sunny_sXe [5.5K]
We can do this with the conservation of momentum. The fact it is elastic means no KE is lost so we don't have to worry about the loss due to sound energy etc.

Firstly, let's calculate the momentum of both objects using p=mv:

Object 1:
p = 0.75 x 8.5 = 6.375 kgm/s

Object 2 (we will make this one negative as it is travelling in the opposite direction):
p = 0.65 x -(7.2) = -4.68 kgm/s

Based on this we know that the momentum is going to be in the direction of object one, and will be 6.375-4.68=1.695 kgm/s

Substituting this into p=mv again:

1.695 = (0.75+0.65) x v
Note I assume here the objects stick together, it doesn't specify - it should!

1.695 = 1.4v
v=1.695/1.4 = 1.2 m/s to the right (to 2sf)
8 0
3 years ago
Which property helps to explain differences in the specific heat capacities of
tatyana61 [14]

Answer:

D. Forces between molecules

Explanation:

Specific heat capacity of water can be defined as the amount of heat a gram of water must lose or absorb in order to change its temperature by a degree Celsius. It is measured in Joules per kilogram per degree Celsius (J/kg°C). Generally, the specific heat capacity of water is 4.182J/kg°C and is the highest among liquids.

Mathematically, the specific heat capacity of a substance is given by the formula;

c = \frac {Q}{mdt}

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

Cohesion is a property of water and it typically refers to the attraction between molecules of water which holds them together.

In Science, the property which helps to explain differences in the specific heat capacities of two substances is the forces between molecules.

This ultimately implies that, the more closely bonded the atoms of a substance are, the higher or greater would be the substance's specific heat capacity. Thus, it varies for the various states of matter i.e solid, liquid and gas.

4 0
3 years ago
Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope
Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

       

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

       

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

#SPJ1

4 0
2 years ago
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