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faust18 [17]
3 years ago
8

g A 39.0-kilogram child sits on a uniform seesaw of negligible mass; she is 2.60 m from the pivot point (or fulcrum). How far fr

om the pivot point on the other side will her 22.0-kilogram playmate have to sit for the seesaw to be in equilibrium?
Physics
1 answer:
Deffense [45]3 years ago
7 0

Answer:

4.6091 meters

Explanation:

In this problem, to have equilibrium in the seesaw, we need the torque from one child to compensate the torque from the other child.

The torque is calculated by making the product of the mass and the distance to the pivot point.

So, the torque generated by the first child (T1) is equal to:

T1 = m1 * d1 = 39 * 2.6 = 101.4 Nm

To make equilibrium, we need the second child to generate the same torque (T2), so:

T2 = m2 * d2 = 22 * d2 = 101.4

d2 = 101.4 / 22 =  4.6091 m

The second child need to be at 4.6091 meters from the pivot point to be in equilibrium.

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A pendulum is formed by taking a 2.0 kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. it is
Lera25 [3.4K]

Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

\frac{\Delta l}{l_{o}}=\frac{F}{AY} (1)

T=2 \pi \sqrt{\frac{l_{o}}{g}} (2)

Where:

\Delta l=0.05 mm=5(10)^{-5} m is the length the steel wire streches (taking into account 1mm=0.001 m)

l_{o} is the length of the steel wire before being streched

F=mg=(2 kg)(9.8 m/s^{2})=19.6 N is the force due gravity (the weight) acting on the pendulum with mass m=2 kg

A is the transversal area of the wire

Y=2(10)^{11} Pa is the Young modulus for steel

T is the period of the pendulum

g=9.8 m/s^{2} is the acceleration due gravity

Knowing this, let's begin by finding A:

A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4} (3)

Where d=1.1 mm=0.0011 m is the diameter of the wire

A=\pi \frac{(0.0011 m)^{2}}{4} (4)

A=9.5(10)^{-7}m^{2} (5)

Knowing this area we can isolate l_{o} from (1):

l_{o}=\frac{\Delta l AY}{F} (6)

And substitute l_{o} in (2):

T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}} (7)

T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}} (8)

Finally:

T=1.39 s

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3 years ago
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Option c. are large

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Other factor is the fact that water has a really high specific heat. This means that water, and more specifically steam, can aborb and transport more energy. A lower heat capacity would imply the need to boil more of the liquid to obtain the same amount of energy. This combine with the fact that water expands at a large rate when boiling, combine with everything mentioned previously, and you get a liquid with all the characteristics that a efficient turbine requires to work.

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