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ElenaW [278]
3 years ago
10

How much heat transfer is necessary to raise the temperature of a 13.6 kg piece of ice from −20.0ºC to 130ºC? specific heat capa

city of water is 4184 J/kg C and specific latent heat of fusion is 334000 J/kg.
8535360 J
4542400 J
458720900 J
13077760 J
Physics
1 answer:
Talja [164]3 years ago
7 0

Answer:

A: 8535360 J

Explanation:

We are given;

mass; m = 13.6 kg

Initial temperature; T1 = -20° C

Final temperature; T2 = 130° C

Specific heat capacity; c = 4184 J/kg.°C

Specific latent heat of fusion; L = 334000 J/kg.

Since we are dealing with change in temperature, then we will use the heat energy formula for specific heat capacity.

Q = mcΔt

Thus;

Q = 13.6 × 4184 × (130 - (-20))

Q = 13.6 × 4184 × 150

Q = 8535360 J

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Explanation:

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Therefore, magnitude of the potential difference between the two spheres is 38.7 volts.

(2).  Now, formula to calculate the energy stored in the capacitor is as follows.

           E = \frac{1}{2}QV

              = \frac{1}{2} \times 3.30 \times 10^{-9} \times 3.87 V

              = 6.39 \times 10^{-8} J

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Three point charges are placed on the x−y plane: a + 50.0-nC charge at the origin, a −50.0-nC charge on the x axis at 10.0 cm, a
butalik [34]

Answer:

(a) F = 0.00322i - 0.00793j with magnitude |F| = 0.00856N

(b) E = -42846.7 N/C

Explanation:

The diagram attached below explains some parameters.

Parameters given:

Charge Q1 = +50 nC at point (0, 0)

Charge Q2 = -50 nC at point (0.1, 0)

Charge Q3 = +150 nC at point (0.1, 0.08)

* The distances are in meters.

(a) The total electric force on the charge Q3 due to Q1 and Q2 is the vector sum of the forces due to Q1 and Q2. Mathematically,

F = F1 + F2

FORCE DUE TO Q1 i.e. F(Q1, Q3)

We have to find the x and y components.

From the diagram, we can find θ using SOHCAHTOA:

θ = tan⁻¹ (0.08/0.1)

θ = 38.66⁰

The distance between Q1 and Q3 can be found using Pythagoras theorem:

x² = 0.08² + 0.1²

x = 0.128 m

F1 = Fx(Q1, Q3)i + Fy(Q1, Q3)j

F1 = iF(Q1, Q3)cosθ + jF(Q1, Q3)sinθ

F(Q1, Q3) = (k * Q1 * Q3) / r²

k = Coulombs constant

F(Q1, Q3) = (9 * 10⁹ * 50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.128)²

F(Q1, Q3) = 0.00412N

F1 = i0.00412 * cos38.66 + j0. 00412 * sin38.66

F1 = 0.00322i + 0.00257j N

FORCE DUE TO Q2 i.e. F(Q2, Q3)

We have to find the x and y components.

F2 = Fx(Q2, Q3)i + Fy(Q2, Q3)j

F2 = iF(Q2, Q3)cos90 + jF(Q2, Q3)cos0

F(Q2, Q3) = (k * Q2 * Q3) / r²

F(Q2, Q3) = (9 * 10⁹ * -50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.08)²

F(Q2, Q3) = -0.0105N

F2 = -i0.0105 * cos90 - j0.0105 * cos0

F2 = - 0.0105j N

Hence, the total force will be

F = F1 + F2

F = 0.00322i + 0.00257j - 0.0105j

F = 0.00322i - 0.00793j N

The magnitude of this force is:

|F| = √(0.00322² + (-0.00793²)

|F| = 0.00856N

(b) The electric field at charge Q3 is the sum of the electric fields due to Q1 and Q2:

E = E1 + E2

E1, electric field due to Q1 = kQ1/r²

E1 = (9 * 10⁹ * 50 * 10⁻⁹) / (0.128²)

E1 = 27465.8 N/C

E2, electric field due to Q2 = (9 * 10⁹ * -50 * 10⁻⁹) / (0.08²)

E1 = -70312.5N/C

The total electric field:

E = E1 + E2

E = 27465.8 - 70312.5

E = -42846.7 N/C

3 0
3 years ago
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