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Citrus2011 [14]
3 years ago
7

A uniform solid cylinder with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with an angular speed of 33.4

rpm. What is its kinetic energy?
Physics
1 answer:
Bas_tet [7]3 years ago
8 0

Answer:

0.092 J

Explanation:

Rotational kinetic energy is:

RE = ½ Iω²

where I is the moment of inertia and ω is the angular speed.

For a solid cylinder, I = ½ mr².

RE = ½ (½ mr²) ω²

RE = ¼ mr²ω²

Convert ω from rpm to rad/s.

33.4 rev/min × (2π rad/rev) × (1 min / 60 s) = 3.50 rad/s

Plug in values and solve:

RE = ¼ (3.0 kg) (0.10 m)² (3.50 rad/s)²

RE = 0.092 J

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A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of t
mash [69]

Answer:

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02 \frac{m}{s^{2}}

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

K=U

\frac{mv^2}{2}=\frac{kx_{max}^2}{2}

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s}

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=

v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}

part 3

Acceleration can be find using Newton's second law:

F=ma

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

-kx=ma

a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s

So between 0 and 4.5 cm we have half a period:

t=\frac{T}{2}=0.66s

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Explanation:

3 and 4G networks, Bluetooth, and Wi-Fi technologies. my opinions

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3 years ago
1. seesaw
Sergeeva-Olga [200]

Answer:

1  seesaw = first class lever

2. pencil and sharpener = wheel and axle

3. bottle opener = second class lever

4. forearm = third class lever

5. nail = wedge

Explanation:

The given objects are classic examples of simple machines.

Lever: There are 3 types of levers and they depend on where the load, applied force, and fulcrum are.

First class levers have the fulcrum in between the load and the applied force. In other words, the load and applied force are at opposite ends. The seesaw is a good example of this. Other examples would be, pliers, scissors, and the like.

Second class levers have the fulcrum and the applied force at the opposite ends. So in this case, the load is found in between. Examples of this would be a bottle opener or a crowbar.

For third class levers, the load and the fulcrum are at opposite ends and the applied force is in between. The forearm is a good example, so is a stapler, or even a broom.

A wedge is another simple machine. It is thick at one end and it gets thinner towards the other end, or it usually has a sharp end. Other examples of this would be an axe, or a knife.

A sharpener is actually a compound machine. The old-fashioned type of pencil sharpener, the one you crank makes use of a wheel and axle and a wedge. A wheel and axle is usually a machine that makes use of two circular parts; a wheel and a rod that is attached to its center.

4 0
3 years ago
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