Answer: The wheel's average rotational acceleration is -0.4 radians per second squared (rad/s^2)
Explanation: Please see the attachments below
If m has a magnitude greater than 1 the image is larger than the object, and an m with a magnitude less than 1 means the image is smaller than the object. If the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object
Answer:
![3.037037037\times 10^{-16}\ C](https://tex.z-dn.net/?f=3.037037037%5Ctimes%2010%5E%7B-16%7D%5C%20C)
Explanation:
dr= Width = 30 μm
R = Radius = 1.8 cm
Q = ![2.05\times 10^6\times 1.6\times 10^{-19}](https://tex.z-dn.net/?f=2.05%5Ctimes%2010%5E6%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D)
r = 0.5 cm
Area is given by
![A=\pi r^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2)
Differentiating with respect to r
![dA=2\pi rdr](https://tex.z-dn.net/?f=dA%3D2%5Cpi%20rdr)
Surface charge density is given by
![\sigma=\dfrac{Q}{A}\\\Rightarrow \sigma=\dfrac{Q}{\pi r^2}](https://tex.z-dn.net/?f=%5Csigma%3D%5Cdfrac%7BQ%7D%7BA%7D%5C%5C%5CRightarrow%20%5Csigma%3D%5Cdfrac%7BQ%7D%7B%5Cpi%20r%5E2%7D)
Charge is given by
![q=\sigma dA\\\Rightarrow q=\dfrac{Q}{\pi r^2} 2\pi rdr\\\Rightarrow q=\dfrac{2Qrdr}{R^2}\\\Rightarrow q=\dfrac{2\times 2.05\times 10^6\times 1.6\times 10^{-19}\times 0.005\times 30\times 10^{-6}}{0.018^2}\\\Rightarrow q=3.037037037\times 10^{-16}\ C](https://tex.z-dn.net/?f=q%3D%5Csigma%20dA%5C%5C%5CRightarrow%20q%3D%5Cdfrac%7BQ%7D%7B%5Cpi%20r%5E2%7D%202%5Cpi%20rdr%5C%5C%5CRightarrow%20q%3D%5Cdfrac%7B2Qrdr%7D%7BR%5E2%7D%5C%5C%5CRightarrow%20q%3D%5Cdfrac%7B2%5Ctimes%202.05%5Ctimes%2010%5E6%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%200.005%5Ctimes%2030%5Ctimes%2010%5E%7B-6%7D%7D%7B0.018%5E2%7D%5C%5C%5CRightarrow%20q%3D3.037037037%5Ctimes%2010%5E%7B-16%7D%5C%20C)
The charge contained in the ring is ![3.037037037\times 10^{-16}\ C](https://tex.z-dn.net/?f=3.037037037%5Ctimes%2010%5E%7B-16%7D%5C%20C)
The answer should be copper. hope this helps :D
80 joule is momentarily stored in each spring
<em><u>Solution:</u></em>
Given that,
The coil springs on a car's suspension have a value of k = 64000 N/m
When the car strikes a bump the springs briefly compress by 5.0 cm (.05 m)
By compressing the spring, we apply a force over a distance
As a result we have done work on the spring
Doing work means that we have transferred energy to spring in form of elastic potential
Therefore,
k = 64000 N/m
x = 0.05m
<em><u>The elastic potential energy is given as:</u></em>
![PE = \frac{1}{2}kx^2](https://tex.z-dn.net/?f=PE%20%3D%20%5Cfrac%7B1%7D%7B2%7Dkx%5E2)
Where, "k" is the spring constant and "x" is the displacement
![PE = \frac{1}{2} \times 64000 \times 0.05^2\\\\PE = 32000 \times 0.0025\\\\PE = 80](https://tex.z-dn.net/?f=PE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%2064000%20%5Ctimes%200.05%5E2%5C%5C%5C%5CPE%20%3D%2032000%20%5Ctimes%200.0025%5C%5C%5C%5CPE%20%3D%2080)
Thus 80 joule is momentarily stored in each spring