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Art [367]
3 years ago
15

A loop of wire lies flat on a horizontal surface. A bar magnet is held above the center of a loop with south pole pointing down.

Magnet is released.
As magnet approaches the loop of wire, the induced current in this wire is:

a) Clockwise
b) Counterclockwise
c) Zero
Physics
2 answers:
Lina20 [59]3 years ago
4 0

Answer:

A

Explanation:

Current, magnetic field and motion are mutually dependent and perpendicular to one another

cricket20 [7]3 years ago
3 0

Answer:

b) clockwise

Explanation:

The right hand rule states that to determine the direction of the magnetic force on a positive moving charge, point the thumb of the right hand in the direction of the potential v, the fingers in the direction of the magnetic field, B, and a perpendicular to the palm points in the direction of the force, F.

Since the south pole of the bar magnet is pointing downwards, the induced current developed in the wire repels it and moves in the clockwise direction.

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Answer:

F = 51.3°

Explanation:

The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

F = Parallel\ Component\ of\ Weight\ of\ Wagon= WSin\theta\\

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6kg of human blood at a temperature of 65degees celcius is mixed with 4kg of human blood ata temperature of 20 degrees celcius .
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Answer:

T_f=47^{\circ}

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(1) Formula basically tells us that the product of mass and temperature remains constant throughout, so the addition of two products of the two separate blood samples would be equal to the product of final temperature and the total mass of the mixture. Mathematically this means that:

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Using (1) and plugging in the corresponding values, we get the answer as follows:

\begin{gathered} T_f=\frac{(m_1\cdot T_1+m_2T_2)}{(m_1+m_2)}\Rightarrow(1) \\ m_1=6\operatorname{kg} \\ m_2=4\operatorname{kg} \\ T_1=65^{\circ} \\ T_2=20^{\circ} \\ \therefore\rightarrow \\ T_f=\frac{(6kg\cdot65^{\circ}+4\operatorname{kg}\cdot20^{\circ})}{(6kg+4\operatorname{kg})}=\frac{(390+80)}{10}=\frac{470}{10}=47^{\circ} \\ \therefore\rightarrow \\ T_f=47^{\circ} \end{gathered}

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