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shusha [124]
3 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]3 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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What is the action force of a flying bird?
kirill [66]
It is called the reaction force of a bird flying.
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7. A 1000 kg car is rolling down the street at 2.5 m/s. How fast would a 2500 kg car have to
babunello [35]

1 m/s

Explanation:

To solve this question we use the following formula:

momentum = mass × velocity

momentum of the first car = 1000 kg × 2.5 m/s

momentum of the second car = 2500 kg × X m/s

To bring the cars at rest the momentum of the first car have to be equal to the momentul of the second car.

momentum of the first car = momentum of the second car

1000 kg × 25 m/s = 2500 kg × X m/s

X (velocity of the second car) = (1000 × 25) / 2500 = 1 m/s

Learn more about:

momentum

brainly.com/question/13378780

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3 years ago
While playing baseball with your friends your hands begin to sting after you ctach several fast balls.
muminat

Answer:

Explanation:

To stop a ball with high momentum in a small-time imparts a high amount of impact on hands. This is the reason for the stinging of hands.

The momentum of the ball is due to the mass and velocity. To prevent stinging in the hand one needs to lower his hands to increase the time of contact. In this way, the momentum transfer to the hands will be lesser.        

7 0
3 years ago
Babies typically say their first words:
deff fn [24]

Answer:

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3 years ago
A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

Q' = + 42 nC = 42\times 10^{- 9} C

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E} = 708.03 N/C

Now,

Electric field at the mid-point due to charge Q' is given by:

\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E'} = 418.84 N/C

Now,

The net Electric field is given by:

\vec{E_{net}} = \vec{E} - \vec{E'}

\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C

5 0
3 years ago
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