Answer
(MG-24)
I pretty sure this the answer your looking for
The elastic potential energy stored in the stretched spring is 1 J.
<h3>What is Hooke's law?</h3>
Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.
Given that;
Force on the spring = 350 Newton
Distance stretched = 7 centimeters or 0.07 m
Hence;
F = ke
k = F/e = 350 Newton/0.07 m = 5000 N/m
Work done in stretching a spring = 1/2ke^2
= 0.5 × 5000 × (2 × 10^-2)^2 =1 J
Learn more about elastic potential energy: brainly.com/question/156316
Answer:
We multiply 4.24 by 38.0 to get the answer.
Explanation:
Time taken by the static-dissipative shoes to do the same thing is 0.78 ms.
The initial potential difference
is related to the final potential difference
by
![\Delta V=\Delta V_{o} e^{-t / \tau}](https://tex.z-dn.net/?f=%5CDelta%20V%3D%5CDelta%20V_%7Bo%7D%20e%5E%7B-t%20%2F%20%5Ctau%7D)
Our target is to find t, so we rearrange equation (1) for t to be
![\Delta V=\Delta V_{o} e^{-t / RC}\\ \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) = \frac{-t}{RC}\\ t= -RC \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) \tag{2}](https://tex.z-dn.net/?f=%5CDelta%20V%3D%5CDelta%20V_%7Bo%7D%20e%5E%7B-t%20%2F%20RC%7D%5C%5C%20%5Cln%20%5Cleft%20%28%20%5Cfrac%7B%5CDelta%20V%7D%7B%5CDelta%20V_%7Bo%7D%7D%20%5Cright%20%29%20%3D%20%5Cfrac%7B-t%7D%7BRC%7D%5C%5C%20t%3D%20-RC%20%5Cln%20%5Cleft%20%28%20%5Cfrac%7B%5CDelta%20V%7D%7B%5CDelta%20V_%7Bo%7D%7D%20%5Cright%20%29%20%5Ctag%7B2%7D)
The body has a capacitance 150 pF while the foot has a capacitance 80 pF and both are in parallel connection. So, the equivalent capacitance is
![C= 150 \mathrm{~pF} + 80 \mathrm{~pF} = 230 \mathrm{~pF}](https://tex.z-dn.net/?f=C%3D%20150%20%5Cmathrm%7B~pF%7D%20%2B%2080%20%5Cmathrm%7B~pF%7D%20%3D%20230%20%5Cmathrm%7B~pF%7D)
The resistance of the static-dissipative shoes is
, so we use this value to find t as in part (a)
![t&= -RC \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) \\ &= -( 1 \times 10^{6} \mathrm{~\Omega})(230 \times 10^{-12} \mathrm{~F}) \ln \left ( \frac{100\mathrm{~V}}{3000 \mathrm{~V}} \right ) \\ &= 0.78\times 10^{-3} \mathrm{~s}\\ &= {0.78\mathrm{~ms}}](https://tex.z-dn.net/?f=t%26%3D%20-RC%20%5Cln%20%5Cleft%20%28%20%5Cfrac%7B%5CDelta%20V%7D%7B%5CDelta%20V_%7Bo%7D%7D%20%5Cright%20%29%20%5C%5C%20%26%3D%20-%28%201%20%5Ctimes%2010%5E%7B6%7D%20%5Cmathrm%7B~%5COmega%7D%29%28230%20%5Ctimes%2010%5E%7B-12%7D%20%5Cmathrm%7B~F%7D%29%20%5Cln%20%5Cleft%20%28%20%5Cfrac%7B100%5Cmathrm%7B~V%7D%7D%7B3000%20%5Cmathrm%7B~V%7D%7D%20%5Cright%20%29%20%5C%5C%20%26%3D%200.78%5Ctimes%2010%5E%7B-3%7D%20%5Cmathrm%7B~s%7D%5C%5C%20%26%3D%20%7B0.78%5Cmathrm%7B~ms%7D%7D)
Learn more about potential difference here:
brainly.com/question/23716417
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Answer:
density of the ball is 3.33 g/cc
Explanation:
As we know that the density is the ratio of mass and volume
here we know that
mass = 20 g
volume = 6 cubic cm
so we will have
![\rho = \frac{m}{V}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7Bm%7D%7BV%7D)
![\rho = \frac{20}{6} g/cm^3](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7B20%7D%7B6%7D%20g%2Fcm%5E3)
![\rho = 3.33 g/cm^3](https://tex.z-dn.net/?f=%5Crho%20%3D%203.33%20g%2Fcm%5E3)