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mixer [17]
3 years ago
14

1. seesaw

Physics
1 answer:
Sergeeva-Olga [200]3 years ago
4 0

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.

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Before the engines fail, the rocket's altitude at time <em>t</em> is given by

y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

and its velocity is

v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t

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1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

Solve for <em>t</em> to find this time to be

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At this time, the rocket attains a velocity of

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When it's in freefall, the rocket's altitude is given by

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(b) Recall that

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where v_f and v_i denote final and initial velocities, respecitively, a denotes acceleration, and \Delta y the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means y_2 will contain the information we need to find the maximum height.

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v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}

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