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3 years ago

8

What will happen if you place too many washers on the end of a shaded-pole motor shaft? A. The motor will become noisy. B. The r

otor will jam. C. The bearings will become too heavy. D. The shaft will loosen.

2 answers:

garik1379 [7]3 years ago

7 0

B. is the right answer

iren [92.7K]3 years ago

3 0

The answer is letter b, the rotor will jam. It is because if there are too many washers, it will be overcrowded, making the rotor to jam in it, where this will lead the motor to dysfunction or not function properly. It is best not to place too many washers in the end of the shaded pole motor shaft to prevent further complications.

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an electric current 0.75 a passes through a circuit that has a resistance of 175. according to ohm's law, what is the voltage of

Aleksandr [31]

Answer:

131.25

Explanation:

i worked it out on a diffrent sheet so its hard to explain

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3 years ago

If a farsighted person has a near point that is 0.600 mm from the eye, what is the focal length f2f2f_2 of the contact lenses th

Readme [11.4K]

Answer:

0.22mm

Explanation:

A far sighted person is a person suffering from long sightedness i.e such individual can only see far distant object clearly but not near distant object. The defect is corrected using convex lens.

Since convex lens is used, the focal (f) length of the lens is positive and the image distance (v) is also positive.

Using the lens formula,

1/f = 1/u + 1/v

Where u is the object distance = 0.35mm

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1/f = 4.53

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3 years ago

A 20 cm-radius ball is uniformly charged to 71 nC.

artcher [175]

Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

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3 years ago

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Serhud [2]

Answer:

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For the wave in this problem,

f = 10 Hz is the frequency

$\lambda=2 m$ is the wavelength

So the speed is

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liubo4ka [24]

Answer:

A proposed answer to a s scientific problem is a hypothesis.

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