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marishachu [46]
3 years ago
13

Compared to its weight on Earth, the weight of a 10N object will be

Physics
2 answers:
algol [13]3 years ago
7 0

What are we supposed to compare it to ?

And why is it called "a 10N object" ?

Alex Ar [27]3 years ago
5 0

The mass of an object is the quantity of matter it possesses. The important thing to remember about mass is that it is constant, and therefore if you are on Earth, the Moon, Mars, or anywhere, your mass is always going to be the same.

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This is false.

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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.

7 0
3 years ago
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