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crimeas [40]
3 years ago
11

Suppose you have two magnets. Magnet A doesn't have its poles labeled, but Magnet B does have a clearly labeled north and south

pole. If these two are brought into contact with one another, which of the following could you expect?
A. The side of Magnet A that's attracted to Magnet B's north pole must be Magnet A's north pole.
B. The side of Magnet A that's repelled by Magnet B's south pole must be Magnet A's north pole.
C. The side of Magnet A that's attracted to Magnet B's south pole must be Magnet A's north pole.
D. The side of Magnet A that's repelled by Magnet B's north pole must be Magnet A's south pole
Physics
2 answers:
Svetllana [295]3 years ago
8 0

Like poles of 2 magnets Repel each other and unlike poles attract each other. so the answer would be C. the side of magnet A that's attracted to Magnet B's South pole must be Magnet A's North pole.

Levart [38]3 years ago
4 0
<span>The right answer here would be option C - the side of Magnet A that's attracted to Magnet B's south pole must be Magnet A's north pole. This is due to the magnetic rule of opposites attracting.</span>
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A constant force of 2.5 N to the right acts on a 4.5 kg mass for 0.90 s.
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Answer:

(a) v_f=0.5\frac{m}{s}

(b) v_f=-11\frac{m}{s}

Explanation:

(a) Since a constant external force is applied to the body, it is under an uniformly accelerated motion. Using the following kinematic equation, we calculate the final velocity of the mass  if it is initially at rest(v_0=0):

v_f=v_0+at\\v_f=at(1)

According to Newton's second law:

F=ma\\a=\frac{F}{m}(2)

Replacing (2) in (1):

v_f=\frac{F}{m}t\\v_f=\frac{2.5N}{4.5kg}(0.9s)\\v_f=0.5\frac{m}{s}

(b) In this case we have v_0=-11.5\frac{m}{s}. So, we use the final velocity equation:

v_f=v_0+at\\v_f=v_0+\frac{F}{m}t\\v_f=-11.5\frac{m}{s}+\frac{2.5N}{4.5kg}(0.9s)\\v_f=-11\frac{m}{s}

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3 years ago
A carpenter is driving a 15.0-g steel nail into a board. His 1.00-kg hammer is moving at 8.50 m/s when it strikes the nail. Half
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Answer: The increase in temperature of the nail after the three blows is 8.0636 Kelvins. The correct option is (d).

Explanation:

Kinetic energy of the hammer ,K.E.=

\frac{1}{2}mv^2=\frac{1}{2}1.00 kg\times (8.50 m/s)^2=36.125 J

Half of the kinetic energy of the hammer is transformed into heat in the nail.

Energy transferred to the nail in one blow =

\frac{1}{2}K.E.=\frac{1}{2}\times 36.125 J=18.0625 J

Total energy transferred after 3 blows,Q =3\times 18.0625 J=54.1875 J

Mass of the nail = 15 g = 0.015 kg

Change in temperature =\Delta T

Specif heat of the steel = c = 448 J/kg K

Q=mc\Delta T

54.1875 J=0.015 kg\times 448 J/kg K\times \Delta T

\Delta T=8.0636 K\approx 8.1 K

The increase in temperature of the nail after the three blows is 8.1  Kelvins.Hence, correct option is (d).

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