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Arisa [49]
3 years ago
6

A coil of wire with 60 turns lies on the plane of the page, and a uniform magnetic field points into the page with a strength of

1.30 T. The coil has an initial area of 0.790 m2 and is then elongated so that its area decreases to 0.370 m2 in 0.80 s. What is the magnitude of the average induced emf in the coil while its area is changing
Physics
1 answer:
damaskus [11]3 years ago
3 0

Answer: -40.95volts

Explanation:

The emf induced when the area of a coil changes in a magnetic field is given by:

E = -NB(∆A/∆t)

B =1.30T, N=60 turns,

∆A =0.790- 0.370 = 0.42

∆t = 0.8s

E = - 1.3 ×60 × (0.42/0.8)

E = - 40.95volts

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3.A golf ball is hit of it’s tee, 200m down the fairway.
Alinara [238K]

<u>Answers:</u>


3. The diagrams showing the forces acting on the golf ball are in the figure attached. Let’s have a detailed look:


a) Here the ball is under 2 forces:


F1 which is called The Normal force and is perpendicular to the surface of the tee where the ball rests


-F1, related to the gravity force, to the weight of the ball, and has the same magnitude, but the opposite direction (That’s why it has a negative sign).



In this case the sum of the forces is 0



b) Here we have again forces F1 and –F1, but in this very moment the club strikes the ball and we have:


F2, the force of the strike





c) While the ball is in its flight, it is under the following forces:


F1, the force of the lift through the air


-F2, the gravity force

-F3, the force of air resistance, also called drag


F4, the tangencial force of the ball flight




4. Here are the sizes and directions of the resultant forces:


i)Two forces of the same magnitude or size are applied to this block, but in opposite directions (in the x-axis). This is expressed as:



F=-10N+10N=0 The resulting force is zero



ii) Two forces of different size and opposite directions in the y-axis are applied to this block. The sum of the forces is:



F=30N-40N=-10N This means the resulting force is 10N applied downwards



iii) In this case the only force applied to the block is -5N applied downwards



iv) Here there are four forces applied to the block.  

In the y-axis we have to forces of the same size but opposite directions:



F1=10N-10N=0 This means the applied force in the y-axis is zero



In the x-axis we have two forces of different size and opposite directions:



F2=-15N+10N=-5N This means the resulting force is applied to the –x-side



4 0
3 years ago
A photon with wavelength λ = 0.0830 nm is incident on an electron that is initially at rest. if the photon scatters in the backw
QveST [7]
Compton scattering equation of wavelengths 
λ'-λ = h/mec (1 - CosФ)
λ' = λ + h/mec (1 - cos 180°)
= ( 0.0830nm) + (6.626 × 10⁻³⁴ J.s)/ (9.1 × 10⁻³¹ kg)(3.0 × 10⁸ m/s
= 0.0830nm
The momentum of electron is
P photon λ = Pe + P phpton λ'
Pe = h/λ - ( -h/λ') = h(λ' + λ)/λλ'
= (6.626 × 10⁻³⁴ J.s)(( 0.08785nm) +( 0.0883nm)/0.08785nm)( 0.083nm)
1.55 × 10⁻²³ kg.m/s.
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Explanation:

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The op amp in this circuit is ideal. R3 has a maximum value of 100 kΩ and σ is restricted to the range of 0.2 ≤ σ ≤ 1.0. a. Calc
Firlakuza [10]

I have attached the circuit image missing in the question.

Answer:

A) The range of vo is; -6.6V≤ vo ≤-1V

B) σ = 0.1861

Explanation:

A) First of all, Let VΔ be the voltage from the potentiometer contact to the ground.

Thus; [(0 - vg)/(2000)] +[(0 - vΔ)/(50,000)] = 0

So, [(- vg)/(2000)] +[(- vΔ)/(50,000)] = 0

Simplifying further; -25 vg - vΔ = 0

From the question, vg = 40mV = 0.04 V

So - 25(0.04) = vΔ

So: vΔ = - 1 V

Now, [vΔ/(σRΔ)] + [(vΔ - 0)/(50,000)] + [(vΔ - vo)/((1 - σ)RΔ))] = 0

So, multiplying each term by RΔ to get; [vΔ/(σ)] + [(vΔ x RΔ)/(50,000)] + [(vΔ - vo)/((1 - σ))] = 0

So RΔ = 100kΩ or 100,000Ω from the question.

So, substituting for RΔ, we get,

[vΔ/(σ)] + [2vΔ] + [(vΔ - vo)/((1 - σ))] = 0

Let's put the value of - 1 for vΔ as gotten before.

So, ( - 1/σ) - 2 + [(-1 - vo)/(1 - σ)] = 0

Now let's make vo the subject of the equation to get;

-1 - vo = (1 - σ)[2 + (1/σ)]

-1 - vo = 2 - 2σ + (1/σ) - 1

-vo = 1 + 2 - 2σ + (1/σ) - 1

-vo = 2 - 2σ + (1/σ)

vo = - 1 (2 - 2σ + (1/σ))

When σ = 0.2; vo = - 1(2 - 0.4 + 5) =

- 1 x 6.6 = - 6.6V

Also when σ = 1;

vo = - 1(2 - 2 + 1) = - 1V

Therefore, the range of vo is;

- 6.6V ≤ vo ≤ - 1V

B) it will saturate at vo = - 7V

So, from;

vo = - 1 (2 - 2σ + (1/σ))

-7 = - 1 (2 - 2σ + (1/σ))

Divide both sides by (-1)

7 = (2 - 2σ + (1/σ))

Now, subtract 2 from both sides to get; 5 = - 2σ + (1/σ)

Multiply each term by α to get;

5σ = - 2σ^(2) + 1

So 2σ^(2) + 5σ - 1 = 0

Solving simultaneously and picking the positive value , we get σ to be approximately 0.1861

8 0
3 years ago
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