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shusha [124]
2 years ago
14

A common pickup for an electric guitar consists of a coil of wire around a small permanent magnet, as described in Figure 25.5.

Why will this type of pickup fail with nylon strings?
Physics
1 answer:
Orlov [11]2 years ago
6 0
Guitar pickups are tiny coils with magnets inside them and they work via  the principal of electromagnetic inductance. The metal strings vibrate within the magnetic field of the pickup which generates an electrical current. Since nylon is not a metal, it will not cause <span>fluctuations in that magnetic field. Hope this answer helps.</span>
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Any child is pushing a shopping cart at a speed of 1.5 m/s.how long will it take this child to push the cart down the aisle with
NARA [144]
1.5 m/s is the velocity. 9.3 m is the length of aisle, over which Distance will be covered. Time is demanded in which the child will move the cart over the aisle with 1.5 m/s. v=S/t and, t=S/v Put values, t=9.3/1.5=6.2 s
7 0
2 years ago
Your little sister is building a radio from scratch. Plans call for a 500 μH inductor wound on a cardboard tube. She brings you
gayaneshka [121]

Answer:

N = 195 turns

Explanation:

The inductance of the inductor, L = 500 μH = 500 * 10⁻⁶H

The length of the tube, l = 12 cm = 0.12 m

The diameter of the tube, d = 4 cm = 0.04 m

Radius, r = 0.04/2 = 0.02 m

Area of the tube, A = πr² = 0.02²π = 0.0004π m²

\mu_{0} = 4\pi * 10^{-7}

The inductance of a solenoid is given by:

L = \frac{\mu_{0}N^{2} A }{l}

500 * 10^{-6} = \frac{4\pi *10^{-7}  N^{2} *4\pi  *10^{-4}  }{0.12}\\500 * 10^{-6} = 0.00000001316N^{2} \\N^{2} = \frac{500 * 10^{-6}}{0.00000001316}\\N^{2} = 37995.44\\N = \sqrt{37995.44} \\N = 194.92 turns

8 0
2 years ago
You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac
dusya [7]

Answer:

20.62361 rad/s

489.81804 J

Explanation:

I_i = Initial moment of inertia = 9.3 kgm²

I_f = Final moment of inertia = 5.1 kgm²

\omega_i = Initial angular speed = 1.8 rev/s

\omega_f = Final angular speed

As the angular momentum of the system is conserved

I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J

The change in kinetic energy of the system is 489.81804 J

As the work was done to move the weight in there was an increase in kinetic energy

6 0
3 years ago
If you wanted to learn more about what it was like to work in a family of scientists, whose life would you research?
Schach [20]

Answer:

Stephen hawking if his family were scientists

6 0
3 years ago
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Based on what you have learned from this unit, construct a tri-fold brochure instructing a new freshman as to the best way to le
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<span>A tri-fold brochure has two parallel folds, splitting the brochure into three sections. Even when printed on low-weight paper, tri-folds can stand up easily, which makes them a great choice for displaying at conventions. You can fold both folds inwards so that the left and right sections of the brochure sit on top of one another, or you can have one fold inwards and the other outwards, to create an accordion effect, which looks very attractive.</span>
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3 years ago
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