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Mrrafil [7]
3 years ago
8

The magnetic flux that passes through one turn of a 8-turn coil of wire changes to 5.0 Wb from 8.0 Wb in a time of 0.098 s. The

average induced current in the coil is 140 A. What is the resistance of the wire
Physics
1 answer:
Advocard [28]3 years ago
7 0

Answer:

<u>Resistance</u><u> </u><u>is</u><u> </u><u>1</u><u>.</u><u>7</u><u>5</u><u> </u><u>ohms</u>

Explanation:

Magnetic flux:

{ \phi{ = NBA}}

N is number of turns, N = 8

B is magnetic flux

A is area of projection.

From faradays law:

E =  -  \frac{ \triangle \phi}{t}

where E is the Electro motive force.

But E = IR

where I is current and R is resistance:

IR =  \frac{( \phi_{1}  -  \phi _{2}) }{t}  \\  \\ 140 \times R =  \frac{8 \times (8 - 5)}{0.098}  \\  \\ R =  \frac{24}{0.098 \times 140}  \\  \\ resistance = 1.75  \: ohms

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Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.8×106N , one an angle 11degrees west of north an
sp2606 [1]

Answer: 2,9 ×10¹⁰J.

Explanation:

1. The <em><u>work </u></em>is a measure of energy and is calculated as the product of the displacement times the parallel force to such displacement.

That means that the only components of the force that contribute to work are those that result parallel to the displacement.

2. Since it is given that the <em>two tugboats "pull the tanker a distance 0.83km toward the north"</em>, that is the displacement, and you have to calculate the net force toward the north.

3. <u>Tugboat #1</u>.

a) Force magnitude: F₁ = 1.8×10⁶N

b) Angle: α = 11° West of North

c) North component of the force F₁: Fy₁ = F₁cos(α) =  1.8×10⁶N  × cos(11°) = 1.77×10⁶N

4. <u>Tugboat #2</u>:

a) Force magnitude: F₂ = 1.8×10⁶N

b) Angle:  = 11° East of North

c) North component of the force F₂: Fy₂ = F₂cos(β) =  1.8×10⁶N  × cos(11°) = 1.77×10⁶N =

5. <u>Total net force, Fn</u>:

Fn = Fy₁ + Fy₂ = 1.77×10⁶N + 1.77×10⁶N = 3.54×10⁶N

6. <u>Work, W</u>:

Displacement, d = 0.83 km = 8,300 m

W = Fn×d = 3.54×10⁶N×8,300m = 29,000 ×10⁶J = 2,9 ×10¹⁰J

The answer is rounded to two significant figures because both data, Force and displacement, have two significant figures.

7 0
3 years ago
In a certain process, the energy of the system decreases by 250 kJkJ. The process involves 480 kJkJ of work done on the system.
BigorU [14]

Answer:

-730KJ

Explanation:

According to the first law of thermodynamics;

Let the total energy of the system be ∆E

Let heat be q and let work the w

Since the energy decreases ∆E is negative

Since work is done on the system w= positive

So;

- 250 = 480 + q

q = -250 - 480

q=-730KJ

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3 years ago
Two electric charges qa = 1. 0 μc and qb = - 2. 0 μc are located 0. 50 m apart. how much work is needed to move the charges apar
melisa1 [442]

The total work done of 0.018 joules is needed to move the charges apart and double the distance between them.

We have two electric charges q(A) = 1μc and q(B) = -2μc kept at a distance 0.5 meter apart.

We have to calculate  much work is needed to move the charges apart and double the distance between them.

<h3>What s the formula to calculate the Potential Energy of a system of two charges (say 'q' and 'Q') separated by a distance 'r' ?</h3>

The potential energy of the system of two charges separated by a distance is given by -

U = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}

In order to solve this question, it is important to remember the work - energy theorem which states -

"The change in the energy of the body is equal to work done on it"

Hence, using this work -energy theorem in the question given to us we get -

U_{f} -U_{i} =W_{net}

In our case -

U_{f}  = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r}\\\\U_{i} =   \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W=\frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r} - \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W = \frac{qQ}{4\pi\epsilon_{o}r} (\frac{1}{2} -1)\\\\W = 9\times 10^{9}\times \frac{1 \times 10^{-6} \times 2\times10^{-6} }{0.5} \times \frac{-1}{2}

W = 0.018 joules

Hence, the total work done should be 0.018 joules.

To solve more question on potential energy, visit the link below -

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4 0
2 years ago
Did the displacement at this point reach its maximum of 2 mm before or after the interval of time when the displacement was a co
Igoryamba

Answer:

a. before

Explanation:

Did the displacement at this point reach its maximum of 2 mm before or after the interval of time when the displacement was a constant 1 mm?

from the graph given from a source. the vertical axis represents the displacement of the graph motion, whilst the horizontal side is representing the time variable of the motion .

displacement is distance in a specific direction.

before the displacement was maximum at 2mm was instant at time=0.04s.

But later was constant at 0.06s at a displacement point of 1mm

4 0
3 years ago
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