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vodka [1.7K]
4 years ago
9

WILL MARK BRAINLIEST PLEASE HELP!! Which class of lever is a seesaw? What are the parts of a seesaw's lever class? Describe each

one. Enter your answer in the space provided.
Physics
1 answer:
kondor19780726 [428]4 years ago
3 0

A seesaw is a class 1 lever

Parts:

Load- The force applied to the lever

Fulcrum- in the middle

Effort- what lifts the load


so in a seesaw with 2 people, whoever is using their feet to push the seesaw up is the effort while the other person is the load. This changes depending on who is pushing the seesaw. the joint/thing between the two people is the fulcrum.



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The radius of a sphere is increasing at a rate of 4 mm/s. how fast is the volume increasing when the diameter is 40 mm?
marin [14]

Using <span>r </span> to represent the radius and <span>t </span> for time, you can write the first rate as:

<span><span><span><span>dr</span><span>dt</span></span>=4<span>mms</span></span> </span>

or

<span><span>r=r<span>(t)</span>=4t</span> </span>

The formula for a solid sphere's volume is:

<span><span>V=V<span>(r)</span>=<span>43</span>π<span>r3</span></span> </span>

When you take the derivative of both sides with respect to time...

<span><span><span><span>dV</span><span>dt</span></span>=<span>43</span>π<span>(3<span>r2</span>)</span><span>(<span><span>dr</span><span>dt</span></span>)</span></span> </span>

...remember the Chain Rule for implicit differentiation. The general format for this is:

<span><span><span><span><span>dV<span>(r)</span></span><span>dt</span></span>=<span><span>dV<span>(r)</span></span><span>dr<span>(t)</span></span></span>⋅<span><span>dr<span>(t)</span></span><span>dt</span></span></span> </span>with <span><span>V=V<span>(r)</span></span> </span> and <span><span>r=r<span>(t)</span></span> </span>.</span>

So, when you take the derivative of the volume, it is with respect to its variable <span>r </span> <span><span>(<span><span>dV<span>(r)</span></span><span>dr<span>(t)</span></span></span>)</span> </span>, but we want to do it with respect to <span>t </span> <span><span>(<span><span>dV<span>(r)</span></span><span>dt</span></span>)</span> </span>. Since <span><span>r=r<span>(t)</span></span> </span> and <span><span>r<span>(t)</span></span> </span> is implicitly a function of <span>t </span>, to make the equality work, you have to multiply by the derivative of the function <span><span>r<span>(t)</span></span> </span> with respect to <span>t </span> <span><span>(<span><span>dr<span>(t)</span></span><span>dt</span></span>)</span> </span>as well. That way, you're taking a derivative along a chain of functions, so to speak (<span><span>V→r→t</span> </span>).

Now what you can do is simply plug in what <span>r </span> is (note you were given diameter) and what <span><span><span>dr</span><span>dt</span></span> </span> is, because <span><span><span>dV</span><span>dt</span></span> </span> describes the rate of change of the volume over time, of a sphere.

<span><span><span><span><span>dV</span><span>dt</span></span>=<span>43</span>π<span>(3<span><span>(20mm)</span>2</span>)</span><span>(4<span>mms</span>)</span></span> </span><span><span>=6400π<span><span>mm3</span>s</span></span> </span></span>

Since time just increases, and the radius increases as a function of time, and the volume increases as a function of a constant times the radius cubed, the volume increases faster than the radius increases, so we can't just say the two rates are the same.

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What is the change in internal energy of the system (∆U) in a process in which 10 kJ of heat energy is absorbed by the system an
Trava [24]

Answer:

Explanation:

According to first law of thermodynamics:

∆U= q + w

= 10kj+(-70kJ)

-60kJ

, w = + 70 kJ

(work done on the system is positive)

q = -10kJ ( heat is given out, so negative)

∆U = -10 + (+70) = +60 kJ

Thus, the internal energy of the system decreases by 60 kJ.

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Answer:

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A wave pulse traveling to the right along a thin cord reaches a discontinuity where the rope becomes thicker and heavier. What i
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Answer:

b) the reflected pulse returns inverted while the transmitted pulse is right side up

Explanation:

A wave pulse traveling to the right along a thin cord reaches a discontinuity where the rope becomes thicker and heavier.

Then, as far as orientation of reflected and transmitted pulses are concerned the reflected pulse returns inverted while the transmitted pulse is right side up.

Hence the correct answer is b.

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