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3 years ago

6

You are looking at your textbook. you can see the textbook because of light doing what process when it hits the textbook's surfa

ce?

1 answer:

Rus_ich [418]3 years ago

5 0

It is because, "Right reflecting" when it hits the textbook's surface, that's why we can see it!!

Hope this helps!

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Which best compares a brush and a commutator in an electric motor?

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The purpose of the brush is supposed to be to lessen the amount of damage or wear placed upon the motor. I hope this helps with your answer choices since you did not post any.

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3 years ago

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Vector A points along the + y axis and a magnitude of 100.0 units. Vector B points at an angle of 60.0 degrees above the +x axis

DedPeter [7]

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vector quantities are resolved into their component form (along the x and y-axis) before adding them. Let us assume that two vectors are

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3 0

3 years ago

You use 350 W of power to move a 7.0 N object 5 m.<br> How long did it take?

PilotLPTM [1.2K]

Answer:

0.1 second

Explanation:

We are given;

Power; P = 350 W

Force; F = 7 N

Distance; d = 5 m

Formula for power is;

P = workdone/time taken

Workdone = F × d

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350 = (7 × 5)/t

t = 35/350

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3 years ago

Would you have been a Federalist or an Anti-Federalist? Why?

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Answer:

Explanation:

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3 years ago

In an experiment designed to measure the strength of a uniform magnetic field produced by a set of coils, electrons are accelera

Arlecino [84]

Answer:

the magnitude of the magnetic field is 8.704 x 10⁻⁴ T

Explanation:

Given;

potential difference, V = 278 V

radius of the circular path, r = 6.46 cm = 0.0646 m

charge of electron, q = 1.60218 × 10⁻¹⁹ C

mass of electron, m = 9.10939 × 10⁻³¹ kg

The magnitude of the magnetic field is given as;

$B = \frac{M_e*v}{q*r}$

where;

B is the magnitude of the magnetic field

$M_e$ is mass of the electron

v is velocity of the electron

r is the radius of the circular path

q is charge of the electron

Determine velocity of the electron from kinetic energy equation;

$K = \frac{1}{2} M_ev^2\\\\Vq = \frac{1}{2} M_ev^2\\\\v^2 = \frac{2qV}{M_e} \\\\v = \sqrt{\frac{2qV}{M_e}} = \sqrt{\frac{2*1.602*10^{-19}*278}{9.109*10^{-31}}} = 9.8886*10^{6} \ m/s$

the magnitude of the magnetic field:

$B = \frac{M_e*v}{q*r} \\\\B = \frac{(9.109*10^{-31})*(9.8886*10^6)}{(1.602*10^{-19})*(0.0646)} = 8.704*10^{-4} \ T$

Therefore, the magnitude of the magnetic field is 8.704 x 10⁻⁴ T

5 0

3 years ago