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Ksivusya [100]
3 years ago
13

An isolated conducting sphere has a 16 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr

ent of 1.0000000 A out of it. How long would it take for the sphere to increase in potential by 1260 V?
Physics
1 answer:
timofeeve [1]3 years ago
5 0

Answer:

It takes 11.20 ms for the sphere to increase in potential by 1260 V

Explanation:

Using the formula q = It where I is current in Ampere, and t is in seconds

q = ( 1.000002 - 1.0000000) t

q = 0.000002t

Voltage on the surface of the sphere V = Kq / r

where K is 9.0 × 10⁹ N².m² / c² and

R = 16 cm = 16 /100 = 0.16m

V = Kq /r

substitute the value into q

V = K(0.000002t) / r

cross multiply

rV = K × 0.000002t

make t subject of the formula

t = 0.16 × 1260 / ( 9×10⁹ × 0.000002)

t = 201.6 / ( 18 × 10³)

t = 0.0112 s

  = 11.20 ms

It takes 11.20 ms for the sphere to increase in potential by 1260 V

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Two Carnot air conditioners, A and B, are removing heat from different rooms. The outside temperature is the same for both rooms
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Answer:

a) Work required for air conditioner A = 354.7 J

b) Work required for air conditioner B = 310.3 J

c) The magnitude of the heat deposited outside for conditioner A = 4684.7 J

d) The magnitude of the heat deposited outside for conditioner B = 4640.3 J

Explanation:

In a carnot air conditioner, it operates like a reverse carnot engine; i.e. it removes heat from the cold reservoir (making it colder) and dumps the heat in the hot reservoir (making it hotter)

For a Carnot air conditioner,

Q꜀ is the heat removed from the colder reservoir = 4330 J for both cases

T꜀ is the temperature of the colder reservoir (temperature of the rooms) = 293 K and 296 K for both cases to be considered.

Qₕ is the heat deposited in the warmer reservoir = ? for both cases

Tₕ is the temperature of the hot reservoir (temperature of outside) = 317 K for both cases.

For Carnot air conditioners,

Qₕ = W + Q꜀ (eqn 1)

And

(Qₕ/Tₕ) - (Q꜀/T꜀) = 0 (eqn 2)

Making Qₕ the subject of formula in eqn 2

Qₕ = Tₕ (Q꜀/T꜀)

Substituting this into eqn 1

Tₕ (Q꜀/T꜀) = W + Q꜀

Q꜀ (Tₕ/T꜀) - Q꜀ = W

Q꜀ [(Tₕ - T꜀)/T꜀ ] = W

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ]

For the air conditioner A,

T꜀ = 293 K, Tₕ = 317 K, Q꜀ = 4330 J, W = ?

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ] = 4330 [ (317 - 293)/293] = 354.7 J

For the air conditioner B,

T꜀ = 296 K, Tₕ = 317 K, Q꜀ = 4330 J, W = ?

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ] = 4330 [ (317 - 296)/296] = 310.3 J

c) Qₕ = W + Q꜀

For conditioner A,

Qₕ = 354.7 + 4330 = 4684.7 J

For conditioner B,

Qₕ = 310.3 + 4330 = 4640.3 J

8 0
3 years ago
A 0. 060-kg tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball initially moving in th
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Final speed of the tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball is 2.964 m/s.

<h3>What is conservation of momentum?</h3>

Momentum of an object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity. By the law of conservation of momentum,

m_1u_1 + m_2u_2 = (m_1+m_2)v

Here, (m) is the mass, (u) is initial velocity before collision, v is final velocity after collision and (subscript 1, and 2) are used for body 1 and 2 respectively. Rewrite the formula for final velocity as,

v=\dfrac{m_1u_1 + m_2u_2}{(m_1+m_2)}

A 0. 060-kg tennis ball, moving with a speed of 5. 82 m/s, has a head-on collision with a 0. 090-kg ball, initially moving in the same direction at a speed of 3.44 m/s. Thus, the initial velocity of the second ball is,

v_{2f}=5.82+3.44+v_{1f}\\v_{2f}=2.38+v_{1f}

Let v1f is the final velocity of first ball. Thus, the initial velocity of the first ball is,

v_{1f}=\dfrac{(0.060)(5.82) + (0.090)(3.44-2.38)}{(0.060)+(0.090)}\\v_{1f}=2.964\rm\; m/s

Thus, final speed of the tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball is 2.964 m/s.

Learn more about the conservation of momentum here;

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Look at a photograph of a fault. notice how the right side appears lower than the left side. this happens when pieces of crust a
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An electron enters the gap between the plates of a capacitor at the center of the gap traveling parallel to theplates at 2.0 x 1
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Answer:

How far will the electron travel beforehitting a plate is 248.125mm

Explanation:

Applying Gauss' law:

Electric Field E = Charge density/epsilon nought

Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12

Therefore E = 1.0 x 10^-6/8.85× 10^-12

E= 1.13×10^5N/C

Force on electron F=qE

Where q=charge of electron=1.6×10^-19C

Therefore F=1.6×10^-19×1.13×10^5

F=1.808×10^-14N

Acceleration on electron a = Force/Mass

Where Mass of electron = 9.10938356 × 10^-31

Therefore a= 1.808×10^-14 /9.11 × 10-31

a= 1.985×10^16m/s^2

Time spent between plate = Distance/Speed

From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2

Therefore Time = 0.01/2×10^6

Time =5×10^-9s

How far the electron would travel S =ut+ at^2/2 where u=0

S= 1.985×10^16×(5×10^-9)^2/2

S=24.8125×10^-2m

S=248.125mm

4 0
3 years ago
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