Answer:
P = 11666.6 W
Explanation:
Given that,
Work done by the motor, W = 3500 kJ
Time, t = 5 min = 300 s
We need to find the power developed by the motor. Power developed is given by :
![P=\dfrac{E}{t}\\\\P=\dfrac{3500\times 10^3}{300}\\\\P=11666.7\ W](https://tex.z-dn.net/?f=P%3D%5Cdfrac%7BE%7D%7Bt%7D%5C%5C%5C%5CP%3D%5Cdfrac%7B3500%5Ctimes%2010%5E3%7D%7B300%7D%5C%5C%5C%5CP%3D11666.7%5C%20W)
So, the required power is 11666.6 W.
- The kinetic energy of the cart at point A is zero and the potential energy is 58,800 J.
- The kinetic energy of the cart at point B is 9,800 J and the potential energy is 49,000 J.
- The velocity of the cart at point B is 9.9 m/s.
<h3>Kinetic and potential energy at point A and B</h3>
At maximum height, velocity is zero and kinetic energy will be zero.
K.E(A) = 0
P.E(A) = mgh = 200 x 9.8 x 30 = 58,800 J
K.E(B) = P.E(A) - P.E(B)
K.E(B) = 58,800 J - (200 x 9.8 x 25)
K.E(B) = 58,800 J - 49,000 J
K.E(B) = 9,800 J
<h3>Velocity at point B</h3>
K.E = ¹/₂mv²
v² = 2K.E/m
v² = (2 x 9800)/(200)
v² = 98
v = √98
v = 9.9 m/s
Learn more about kinetic energy here: brainly.com/question/25959744
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4.33x50
10^5 would be 50 so 50/25
Answer:
a. E = 122.4 N/C
b. E = 58.2 N/C
c. E = 0
Explanation:
The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.
In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.
A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.
![E2\pi rh = \frac{\lambda V}{\epsilon_0} = \frac{\lambda \pi (b^2 - a^2)h}{\epsilon_0}\\E2\pi (1.97)h = \frac{(5.3\times 10^{-9})\pi(1.67^2 - 1.41^2)h}{\epsilon_0}\\E = 122.4~N/C](https://tex.z-dn.net/?f=E2%5Cpi%20rh%20%3D%20%5Cfrac%7B%5Clambda%20V%7D%7B%5Cepsilon_0%7D%20%3D%20%5Cfrac%7B%5Clambda%20%5Cpi%20%28b%5E2%20-%20a%5E2%29h%7D%7B%5Cepsilon_0%7D%5C%5CE2%5Cpi%20%281.97%29h%20%3D%20%5Cfrac%7B%285.3%5Ctimes%2010%5E%7B-9%7D%29%5Cpi%281.67%5E2%20-%201.41%5E2%29h%7D%7B%5Cepsilon_0%7D%5C%5CE%20%3D%20122.4~N%2FC)
B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.
![E2\pi rh = \frac{\lambda V_{enc}}{\epsilon_0} = \frac{\lambda \pi (r^2 - a^2}h{\epsilon_0}\\E2\pi (1.51)h = \frac{5.3\times 10^{-9})\pi(1.51^2 - 1.41^2)h}{\epsilon_0}\\E = 58.2~N/C](https://tex.z-dn.net/?f=E2%5Cpi%20rh%20%3D%20%5Cfrac%7B%5Clambda%20V_%7Benc%7D%7D%7B%5Cepsilon_0%7D%20%3D%20%5Cfrac%7B%5Clambda%20%5Cpi%20%28r%5E2%20-%20a%5E2%7Dh%7B%5Cepsilon_0%7D%5C%5CE2%5Cpi%20%281.51%29h%20%3D%20%5Cfrac%7B5.3%5Ctimes%2010%5E%7B-9%7D%29%5Cpi%281.51%5E2%20-%201.41%5E2%29h%7D%7B%5Cepsilon_0%7D%5C%5CE%20%3D%2058.2~N%2FC)
C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.