The equation Q=CV (Charge = product of Capacitance and potential difference) tells us that the maximum charge that can be stored on a capacitor is equal to the product of it's capacitance and the potential difference across it. In this case the potential difference across the capacitor will be 12.0V (assuming circuit resistance is negligable) and it has a capacitance of 18.0μf or 18.0x10^-6f, therefore charge equals (18.0x10^-6)x12=2.16x10^-4C (Coulombs).
Answer:
2.64 x 10⁻⁶T
Explanation:
The magnitude of the magnetic field produced by a long straight wire carrying current is given by Biot-Savart law as follows: "The magnetic field strength is directly proportional to the current on the wire and inversely proportional to the distance from the wire". This can be written mathematically as;
B = (μ₀ I) / (2π r) ----------------(i)
B is magnetic field
I is current through the wire
r is the distance from the wire
μ₀ is the magnetic constant = 4π x 10⁻⁷Hm⁻¹
From the question;
I = 0.7A
r = 0.053m
Substitute these values into equation (i) as follows;
B = (4π x 10⁻⁷ x 0.7) / (2π x 0.053)
B = 2.64 x 10⁻⁶T
Therefore the approximate magnitude of the magnetic field at that location is 2.64 x 10⁻⁶T
Given data:
* The mass of the baseball is 0.31 kg.
* The length of the string is 0.51 m.
* The maximum tension in the string is 7.5 N.
Solution:
The centripetal force acting on the ball at the top of the loop is,
![\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20T%2Bmg%3D%5Cfrac%7Bmv%5E2%7D%7BL%7D_%7B%7D%20%5C%5C%20v%5E2%3D%5Cfrac%7BL%28T%2Bmg%29%7D%7Bm%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7BL%28T%2Bmg%29%7D%7Bm%7D%7D%20%5Cend%7Bgathered%7D)
For the maximum velocity of the ball at the top of the vertical circular motion,
![v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}](https://tex.z-dn.net/?f=v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7BL%28T_%7B%5Cmax%20%7D%2Bmg%29%7D%7Bm%7D%7D)
where g is the acceleration due to gravity,
Substituting the known values,
![\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B0.51%287.5_%7B%7D%2B0.31%5Ctimes9.8%29%7D%7B0.31%7D%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B0.51%2810.538%29%7D%7B0.31%7D%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B17.34%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D4.16%5Ctext%7B%20m%2Fs%7D%20%5Cend%7Bgathered%7D)
Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.
