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vichka [17]
3 years ago
15

Two campers dock a canoe. One camper has a mass of 100.0 kg and moves forward at 3.0 m/s as he leaves the canoe to step onto the

dock. With what speed do the canoe and other camper move if their combined mass is 175.0 kg
Physics
1 answer:
Solnce55 [7]3 years ago
4 0

Answer:

The combined speed of camper and canoe is 1.71 m/s.

Explanation:

Given that,

Mass of camper 1, m = 100 kg

Speed of camper 1, v = 3 m/s

The combined mass of another camper and canoe is, M = 175 kg

We need to find the combined speed of camper and canoe. According to the conservation of linear momentum, the momentum of first camper is equal to linear momentum of the canoe and the second camper.

mv=MV\\\\V=\dfrac{mv}{M}\\\\V=\dfrac{100\times 3}{175}\\\\V=1.71\ m/s

So, the combined speed of camper and canoe is 1.71 m/s.

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A meter stick is suspended vertically at a pivot point 22 cm from the top end. It is rotated on the pivot until it is horizontal
suter [353]

Answer:

5.82812 rad/s

Explanation:

L = Length of meter stick = 1 m = 100 cm

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\omega = Angular velocity

Moment of inertia of the system is given by

I=I_c+mr^2\\\Rightarrow I=\frac{mL^2}{12}+mr^2\\\Rightarrow I=\frac{m1^2}{12}+m0.28^2\\\Rightarrow I=m(\frac{1}{12}+0.0784)

As the energy in the system is conserved

mgh=I\frac{\omega^2}{2}\\\Rightarrow mgh=m(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow gh=(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow \omega=\sqrt{\frac{2gh}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=\sqrt{\frac{2\times 9.81\times 0.28}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=5.82812\ rad/s

The maximum angular velocity is 5.82812 rad/s

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