Ah ha ! Very interesting question.
Thought-provoking, even.
You have something that weighs 1 Newton, and you want to know
the situation in which the object would have the greatest mass.
Weight = (mass) x (local gravity)
Mass = (weight) / (local gravity)
Mass = (1 Newton) / (local gravity)
"Local gravity" is the denominator of the fraction, so the fraction
has its greatest value when 'local gravity' is smallest. This is the
clue that gives it away.
If somebody offers you 1 chunk of gold that weighs 1 Newton,
you say to him:
"Fine ! Great ! Golly gee, that's sure generous of you.
But before you start weighing the chunk to give me, I want you
to take your gold and your scale to Pluto, and weigh my chunk
there. And if you don't mind, be quick about it."
The local acceleration of gravity on Pluto is 0.62 m/s² ,
but on Earth, it's 9.81 m/s.
So if he weighs 1 Newton of gold for you on Pluto, its mass will be
1.613 kilograms, and it'll weigh 15.82 Newtons here on Earth.
That's almost 3.6 pounds of gold, worth over $57,000 !
It would be even better if you could convince him to weigh it on
Halley's Comet, or on any asteroid. Wherever he's willing to go
that has the smallest gravity. That's the place where the largest
mass weighs 1 Newton.
I uploaded the answer to
a file hosting. Here's link:
bit.
ly/3gVQKw3
Answer:
2*10^-<em>5</em>
Explanation:
<em>B=</em><em>I</em><em>L</em>
<em>I=</em><em>B</em><em>/</em><em>L</em>
<em>I=</em><em>0</em><em>.</em><em>0</em><em>0</em><em>2</em><em>0</em><em>*</em><em>1</em><em>0</em><em>^</em><em>-</em><em>4</em><em>/</em><em>1</em><em>0</em>
<em>I=</em><em>2</em><em>*</em><em>1</em><em>0</em><em>^</em><em>5</em>
OK what is the hole answer i can help you
Answer:
x₂=0.44m
Explanation:
First, we calculate the length the spring is stretch when the first block is hung from it:
![\Delta x_1=0.50m-0.35m=0.15m](https://tex.z-dn.net/?f=%5CDelta%20x_1%3D0.50m-0.35m%3D0.15m)
Now, since the stretched spring is in equilibrium, we have that the spring restoring force must be equal to the weight of the block:
![k\Delta x_1=m_1g](https://tex.z-dn.net/?f=k%5CDelta%20x_1%3Dm_1g)
Solving for the spring constant k, we get:
![k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(6.3kg)(9.8m/s^{2})}{0.15m}=410\frac{N}{m}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Bm_1g%7D%7B%5CDelta%20x_1%7D%5C%5C%5C%5Ck%3D%5Cfrac%7B%286.3kg%29%289.8m%2Fs%5E%7B2%7D%29%7D%7B0.15m%7D%3D410%5Cfrac%7BN%7D%7Bm%7D)
Next, we use the same relationship, but for the second block, to find the value of the stretched length:
![k\Delta x_2=m_2g\\\\\Delta x_2=\frac{m_2g}{k}\\\\\implies \Delta x_2=\frac{(3.7kg)(9.8m/s^{2})}{410N/m}=0.088m](https://tex.z-dn.net/?f=k%5CDelta%20x_2%3Dm_2g%5C%5C%5C%5C%5CDelta%20x_2%3D%5Cfrac%7Bm_2g%7D%7Bk%7D%5C%5C%5C%5C%5Cimplies%20%5CDelta%20x_2%3D%5Cfrac%7B%283.7kg%29%289.8m%2Fs%5E%7B2%7D%29%7D%7B410N%2Fm%7D%3D0.088m)
Finally, we sum this to the unstretched length to obtain the length of the spring:
![x_2=0.35m+0.088m=0.44m](https://tex.z-dn.net/?f=x_2%3D0.35m%2B0.088m%3D0.44m)
In words, the length of the spring when the second block is hung from it, is 0.44m.