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gladu [14]
3 years ago
15

. A current-carrying gold wire has diameter 0.87 mm. The electric field in the wire is 0.54 V>m. What are (a) the current car

ried by the wire; (b) the potential difference between two points in the wire 7.0 m apart; (c) the resistance of a 7.0-m length of this wire?
Physics
1 answer:
mixas84 [53]3 years ago
6 0

Answer:

a) I = 13.156 A

b) V = 3.78 V

c) R = 0.2873 Ω

Explanation:

The diameter(d) of the wire is 0.87 mm = 0.87 × 10⁻³ m.

The electric field(E) = 0.54 V/m

The cross sectional area(A) = πd²/4 = π × (0.87 × 10⁻³)² /4 = 5.945 × 10⁺⁷ m²

Gold resistivity(ρ) = 2.44 × 10⁻⁸

(a) the current carried by the wire

E = Jρ

Where J is the current density = I/A and I is the current

E = Iρ/A

I = EA/ρ

Substituting values:

I = 0.54 × 5.945 × 10⁺⁷/  2.44 × 10⁻⁸ = 13.156 A

(b) the potential difference between two points in the wire 7.0 m apart

The potential difference(V) = EL

Where L is the length = 7 m

V = EL

Substituting values:

V = 0.54 × 7 = 3.78 V

(c) the resistance of a 7.0-m length of this wire

The resistance(R) = ρL/A

R =  ρL/A

Substituting values:

R = 2.44 × 10⁻⁸ × 7 /  5.945 × 10⁺⁷ = 0.2873 Ω

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svetoff [14.1K]

1)

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

v_f = 24.5 m/s

Explanation:

Part 1)

initial speed of the ball upwards

v_i = 24.5 m/s

so maximum height of the ball is given by

H = \frac{v_i^2}{2g}

H = \frac{24.5^2}{2(9.80)}

H = 30.6 m

Part 2)

As we know that final speed will be zero at maximum height

so we will have

v_f - v_i = at

0 - 24.5 = (-9.8)t

t = 2.5 s

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

v_f = 24.5 m/s

2)

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

y = \frac{1}{2}at^2

4.88 = \frac{1}{2}a(1^2)

a = 9.76 m/s^2

3)

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

a = 90 m/s^2

time taken by the rocket

t = 0.33 min

final speed of the rocket is given as

v_f = v_i + at

v_f = 0 + (90)(0.33)

v_f = 29.7 m/s

4)

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

so here we will have

y = \frac{1}{2}gt^2

y = \frac{1}{2}(9.81)(2.3^2)

y = 25.95 m

Part 2)

Distance traveled by it in horizontal direction is given as

d = v_x t

d = 2.92 \times 2.3

d = 6.72 m

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A light bulb operating at a dc voltage of 120 V has a power rating of 60 W. How much current is flowing through this bulb
densk [106]
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  • Power=P=60W

  • Current=I

\\ \rm\rightarrowtail I=\dfrac{P}{V}=\dfrac{60}{120}=\dfrac{1}{2}=0.5A

8 0
2 years ago
What are the advantages of using the metric system? SELECT ALL THAT APPLY
alexandr402 [8]

D. used by the entire scientific community

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An engine absorbs 1749 J from a hot reservoir and expels 539 J to a cold reservoir in each cycle. a. What is the engine’s effici
Masja [62]

Answer:

a)η = 69.18 %

b)W= 1210 J

c)P=3967.21 W

Explanation:

Given that

Q₁ = 1749 J

Q₂ = 539  J

From first law of thermodynamics

Q₁   = Q₂ +W

W=Work out put

Q₂=Heat rejected to the cold reservoir

Q₁ =heat absorb by hot  reservoir

W= Q₁- Q₂

W= 1210 J

The efficiency given as

\eta=\dfrac{W}{Q_1}

\eta=\dfrac{1210}{1749}

\eta=0.6918

η = 69.18 %

We know that rate of work done is known as power

P=\dfrac{W}{t}

P=\dfrac{1210}{0.305}\ W

P=3967.21 W

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