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bulgar [2K]
3 years ago
6

Which of the following factors will increase the current in a stationary wire loop?

Physics
2 answers:
il63 [147K]3 years ago
7 0

Answer:

D) All of the above

Explanation:

As we know by Faraday's law of electromagnetic induction

we will have

EMF = \frac{d\phi}{dt}

here we have

\phi = B.A

so we have

EMF = \frac{d(B.A)}{dt}

now current in the loop is given as

i = \frac{EMF}{R}

i = \frac{1}{R}(\frac{d(B.A)}{dt})

now this current will increase by changing the rate of flux which will change by either increasing or decreasing the speed of magnet

Also we can change by changing the field strength

So here correct answers are

A. Increasing the speed of an approaching magnet

B. Decreasing the speed of a receding magnet

C. Increasing the strength of a stationary magnet

kogti [31]3 years ago
6 0
<span>A. Increasing the speed of an approaching magnet</span>
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Par 1/2
BaLLatris [955]

part 1

mass = ρ x V

mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg

PE (potential energy)= mgh

PE = 6608.2 kg x 9.81 x 403

PE = 2.61 x 10⁷ J

part 2

megaton of TNT (Mt) =4.2 x 10¹⁵ J

convert PE to Mt:

2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt

4 0
3 years ago
Lightning flashes one mile (1609 m) away from you. How much time does it take the light to travel that distance?
hichkok12 [17]

Answer:

t=5.36\times 10^{-6}\ s

Explanation:

Given that,

Lightning flashes one mile (1609 m) away from you.

We need to find the time it take the light to travel that distance. Let the time be t. We know that,

speed = distance/time

t=\dfrac{d}{v}\\\\t=\dfrac{1609}{3\cdot10^{8}}\\\\=5.36\times 10^{-6}\ s

So, the required time is 5.36\times 10^{-6}\ s

7 0
3 years ago
Sal jumps into the air with an initial velocity of 1.1 m/s. How high will he go before he comes to
sammy [17]
1.1 m/s hope this helps.
3 0
2 years ago
A 68.0-kg person jumps from rest off a 2.20-m-high tower straight down into the water. Neglect air resistance during the descent
Margarita [4]

Answer:

F= 1333.767\,N

Explanation:

The velocity of the swimmer just before touching the water is:

v = -\sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.20\,m)}

v \approx 6.569\,\frac{m}{s}

The average force exerted on the diver by the water is determined by the use of the Principle of Energy Conservation and the Work-Energy Theorem:

(68\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (2.20\,m) +\frac{1}{2}\cdot (68\,kg)\cdot (6.569\,\frac{m}{s} )^{2}-F\cdot(2.20\,m) = 0\,J

F= 1333.767\,N

6 0
3 years ago
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3 years ago
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