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3 years ago

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Two boys wearing roller skates are standing on a smooth surface with the palms of their hands touching and their arms bent, as s

hown above. if boy x pushes by straightening his arms out while boy y holds his arms in the oringinal position, what is the motion of the two boys

1 answer:

vlada-n [284]3 years ago

8 0

Boy X and Boy Y both move backward in opposite directions.

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The __________ states that the firing rate of neurons in the auditory nerve matches the frequency of the sound wave.

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C. Volley principle (FREQUENCY MATCHING)

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Consider a solenoid of length L, N windings, and radius b (L is much longer than b A current I). is flowing through the wire. If

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Answer:

The magnetic field inside the solenoid would decrease by a factor of 2.

Explanation:

The magnetic field, B, of a solenoid of length L, N windings, and radius b with a current, I, flowing through it is given as:

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If the length of the solenoid is doubled, 2L,the magnetic field becomes:

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B2 = ½ B

The magnetic field will decrease by a factor of 2.

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3 years ago

Before beginning a long trip on a hot day, a driver inflates anautomobile tire to a gauge pressure of 2.70 atm at 300 K. At the

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Answer:

The value is the temperature of the air inside the tire $T_{2} =$ 340.54 K

% of the original mass of air in the tire should be released 99.706 %

Explanation:

Initial gauge pressure = 2.7 atm

Absolute pressure at inlet $P_{1}$ = 2.7 + 1 = 3.7 atm

Absolute pressure at outlet $P_{2}$ = 3.2 + 1 = 4.2 atm

Temperature at inlet $T_{1}$ = 300 K

(a) Volume of the system is constant so pressure is directly proportional to the temperature.

$\frac{T_{2} }{T_{1} } = \frac{P_{2} }{P_{1} }$

$\frac{T_{2} }{300} = \frac{4.2}{3.7}$

$T_{2} =$ 340.54 K

This is the value is the temperature of the air inside the tire

(b). Since volume of the tyre is constant & pressure reaches the original value.

From ideal gas equation P V = m R T

Since P , V & R is constant. So

m T = constant

$m_{1} T_{1} = m_{2} T_{2}$

$\frac{m_{2} }{m_{1} } = \frac{T_{1} }{T_{2} }$

$\frac{m_{2} }{m_{1} } = \frac{300}{354.54}$

$\frac{m_{2} }{m_{1} } =0.00294$

value of the original mass of air in the tire should be released is $\frac{m_{2} - m_{1}}{m_{1}}$

$\frac{0.00294-1}{1}$

⇒ -0.99706

% of the original mass of air in the tire should be released 99.706 %.

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3 years ago