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2 years ago

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Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon

has an arm length (hand to shoulder) of 0.60 m. We can model its motion as that of a point mass swinging at the end of a 0.60-m-long, massless rod. At the lowest point of its swing, the gibbon is moving at 3.2 m/s .
What upward force must a branch provide to support the swinging gibbon?
Express your answer to two significant figures and include the appropriate units.

1 answer:

aleksklad [387]2 years ago

5 0

Answer:

230 N

Explanation:

At the lowest position , the velocity is maximum hence at this point, maximum support force T is given by the branch.

The swinging motion of the ape on a vertical circular path , will require

a centripetal force in upward direction . This is related to weight as follows

T - mg = m v² / R

R is radius of circular path . m is mass of the ape and velocity is 3.2 m/s

T = mg - mv² / R

T = 8.5 X 9.8 + 8.5 X 3.2² / .60 { R is length of hand of ape. }

T = 83.3 + 145.06

= 228.36

= 230 N ( approximately )

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Answer:

The current is $I = 1 A$

The direction is anti-clockwise

Explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

the length of the conducting rod is $L = 25 \ cm = \frac{25}{100} = 2.5 \ m$

The resistance is $R = 8 \Omega$

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The speed of the rod is $v = 6.0 m/s$

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substituting values we have

$\epsilon = 0.40 * 2.5 * 8$

$\epsilon = 8V$

From ohm law the induced current would be

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substituting values we have

$I = \frac{8}{8 }$

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The direction anticlockwise this because according to lenze law the current due to change in magnetic field will act in the opposite direction of the force causing the magnetic field to change

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3 years ago

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Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

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The rotational inertia of the merry-go-round = 600 kg·m²

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The speed with which the boy approaches the merry-go-round = 5.0 m/s

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Where;

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For the merry go round, we have;

$I_m \cdot \alpha_m = I_m \cdot \dfrac{v_m}{r \cdot t}$

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$\alpha _m$ = The angular acceleration of the merry-go-round

$v _m$ = The linear velocity of the merry-go-round

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For the boy, we have;

$I_b \cdot \alpha_b = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Where;

$I_b$ = The rotational inertia of the boy

$\alpha _b$ = The angular acceleration of the boy

$v _b$ = The linear velocity of the boy

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$I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

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$v_m = \dfrac{m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2 \cdot v_b}{I_m}$

From which we have;

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