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Step2247 [10]
2 years ago
12

A 5.2 cm diameter circular loop of wire is in a 1.35 T magnetic field. The loop is removed from the field in 0.29 sec. Assume th

at the loop is perpendicular to the magnetic field. What is the average induced emf?
Physics
1 answer:
Katyanochek1 [597]2 years ago
3 0

Answer:

9.88 milivolt

Explanation:

Given: diameter d = 5.2 cm

magnetic field B_1 = 1.35 T, final magnetic field B_2 =0 T

t = 0.29 sec.

we know emf =  - dΦ/dt

and flux Φ = BA

A= area

therefore emf ε = -A(B_2-B_1)/Δt

=-\pi(d/2)^2\frac{B_2-B_1}{\Delta t} \\=-\pi(0.052/2)^2\frac{0-1.35}{0.29} \\=98.8\times10^4\\=9.88 mV

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a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it wi
Lelu [443]

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s .The amplitude of the subsequent oscillations 48.13 cm/s

a 1.25 kilogram block is fastened to a spring with a 17.0 newtons per meter spring constant. Given that K is equal to 14 Newtons per meter and mass equals 10.5 kg. The block is then struck with a hammer by a student while it is at rest, giving it a speedo of 46.0 cm for a brief period of time. The required energy provided by the hammer, which is half mv squared, is transformed into potential energy as a result of the succeeding oscillations. This is because we know that energy is still available for consultation. So access the amplitude here from here. He will therefore be equal to and by. Consequently, the Newton's spring constant is 14 and the value is 10.5. The velocity multiplied by 0.49

Speed at X equals 0.35 into amplitude, or vice versa. At this point, the spirit will equal half of K X 1 squared plus half. Due to the fact that this is the overall energy, square is equivalent to half of a K square or an angry square. amplitude is 13 and half case 14 x one is 0.35. calculate that is equal to initial velocities of 49 squares and masses of 10.5. This will be divided in half and start at about 10.5 into the 49-square-minus-14. 13.42 into the entire square in 20.35. dividing by 10.5 and taking the square as a result. 231 6.9 Six centimeters per square second. 10.5 into 49 sq. 14. 2 into a 13.42 square entire. then subtract 10.5 from the result to get the square. So that is 48.13cm/s.

To learn more about oscillations Please click on the given link:

brainly.com/question/26146375

#SPJ4

This is incomplete question Complete Question is:

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s . what are The amplitude of the subsequent oscillations?

4 0
1 year ago
An 800 N man climbs 5 m up a ladder. How much gravitational potential energy does he gain?
Artemon [7]

Answer:

4000J

Explanation:

Given parameters:

Weight of the man  = 800N

Height of ladder  = 5m

Unknown:

Gravitational potential energy gained  = ?

Solution:

The gravitational potential energy is due to the position of a body.

 Gravitational potential energy = weight x height

Now insert the parameters;

 Gravitational potential energy  = 800 x 5  = 4000J

5 0
3 years ago
Find the potential energy associated with a 61-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level. Take the
Natali [406]

Answer:

The potential energy of the hiker is 1.13\times 10^6\ J.

Explanation:

Given that,

Mass of the hiker, m = 61 kg

Height above sea level, h = 1900 m

We need to find the potential energy associated with a 61-kg hiker atop New Hampshire's Mount Washington. The potential energy is given by :

E=mgh

g is the acceleration due to gravity

E=61\ kg\times 9.8\ m/s^2\times 1900\ m\\\\E=1.13\times 10^6\ J

So, the potential energy of the hiker is 1.13\times 10^6\ J. Hence, this is the required solution.

8 0
3 years ago
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