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Artyom0805 [142]
3 years ago
8

A bar magnet is dropped from above and falls through the loop of wire. The north pole of the bar magnet points downward towards

the page as it falls. Which statement is correct?a. The current in the loop always flows in a clockwise direction. b·The current in the loop always flows in a counterclockwise direction. c. The current in the loop flows first in a clockwise, then in a counterclockwise direction. d. The current in the loop flows first in a counterclockwise, then in a clockwise direction. e. No current flows in the loop because both ends of the magnet move through the loop.
Physics
1 answer:
8_murik_8 [283]3 years ago
3 0

Answer:

<em>b. The current in the loop always flows in a counterclockwise direction.</em>

<em></em>

Explanation:

When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.

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Competition in the Los Angeles Flower District results in
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3 years ago
a flashlight has a resistance of 2.4 ohms. what voltage is applied by the batteries if the current in the circuit is 2.5A?
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8 0
4 years ago
Read 2 more answers
Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc
Sergeeva-Olga [200]

Answer:

a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

b) Average power

P(w)= 1062.07 [w]

P(hp)=1.42 [hp]

c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.

Explanation:

First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

Work:

W= F*d= m*g*d

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J]

Converting from Joules to Kcals:

\frac{10620.75}{4186} =2.537 Kcal

Now lets take into account the efficiency of the human body (20%)

2.537 ---> 20%

 x       ---> 100%

x=\frac{2.537*100}{20} =12.685

So the student is consuming 12.685 KCals each time he runs up the stairs.

Now,

1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

\frac{9000Kcals}{12.685Kcals} =709.5 times

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

********************

For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)

Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\

P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]

*****

4 0
3 years ago
Compute the power output (watts) during one minute of treadmill exercise, given the following: Treadmill grade-10% Horizontal sp
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Answer:

c. 981 watts

P=981\ W

Explanation:

Given:

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<u>Now the power can be given by:</u>

P=v.w

P=588.6\times\frac{5}{3} (where grade is the rise of the front edge per 100 m of the horizontal length)

P=981\ W

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