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Oksana_A [137]
3 years ago
6

A heavy boy and a lightweight girl are balanced on a mass-less seesaw. If they both move forward so that they are one-half their

original distance from the pivot point, what will happen to the seesaw? Assume that both people are small enough compared to the length of the seesaw to be thought of as point masses.A. It is impossible to say without knowing the masses.
B. It is impossible to say without knowing the distances.
C. The side the boy is sitting on will tilt downward.
D. Nothing will happen; the seesaw will still be balanced.
E. The side the girl is sitting on will tilt downward.
Physics
2 answers:
Rom4ik [11]3 years ago
4 0

Answer:

b) Nothing will happen,  the sea saw will still be balanced.

Explanation:

b) Nothing will happen,  the sea saw will still be balanced.

Reason:-

When two kids are balanced, the sum of torques on the seesaw will be zero.

if each kid, reduces their distances by half, then the torque of each kid will be half and the sum of torque of each on the seesaw will be zero.

Therefore the seesaw is balanced

Katyanochek1 [597]3 years ago
4 0

Answer:D. Nothing will happen; the seesaw will still be balanced

Explanation: According to the principles of moment which states that for a body to be at equilibrium, the sum of clockwise moments must be equal to the sum of anticlockwise moments.

if the mass-less seesaw was initially balanced at a particular distance, it will still be balanced if the boy and the girl move forward to cover half their initial distances from the pivot.

D perfectly depicts the outcome.

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A negative charge of - 8.0 x 10^-6 C exerts an attractive force of 12 N on a second charge that is
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Answer:

<h2>Magnitude of the second charge is -4.17*10^{-7}C</h2>

Explanation:

According to columbs law;

F = kq1q2/r^{2}

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let q1 = - 8.0 x 10^-6 C

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Cross multiplying

0.03=9*10^{9}*  -8.0*10^{-6} q2\\0.03 = -72*10^{3} q2\\q2 = \frac{0.03}{ -72*10^{3}} \\q2 = -4.17*10^{-7}C

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P_2 = Momentum of the ball after hit

55^{\circ} = Angle ball makes with the horizontal after hitting the pin

\theta = Angle the pin makes with the horizotal after getting hit by the ball

Momentum in the x direction

P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}

Momentum in the y direction

P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}

(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}

The pin's resultant velocity is 14.98\ \text{kg m/s}

P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}

The pin's resultant direction is 45.26^{\circ} below the horizontal or to the right.

4 0
2 years ago
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