Answer:
The work done by gravity during the roll is 490.6 J
Explanation:
The work (W) is:
<em>Where</em>:
F: is the force
d: is the displacement = 20 m
The force is equal to the weight (W) in the x component:
<em>Where:</em>
m: is the mass of the bowling ball = 5 kg
g: is the gravity = 9.81 m/s²
θ: is the degree angle to the horizontal = 30°
Now, we can find the work:
Therefore, the work done by gravity during the roll is 490.6 J.
I hope it helps you!
Answer: 39.8 μC
Explanation:
The magnitude of the electric field generated by a capacitor is given by:
d is the distance between the plates.
For a capacitor, charge Q = CV where C is the capacitance and V is the voltage.
where A is the area of the plate and ε₀ is the absolute permittivity.
substituting, we get
It is given that the magnitude of the electric field that can exist in the capacitor before air breaks down is, E = 3 × 10⁶ N/C.
radius of the plates of the capacitor, r = 69 cm = 0.69 m
Area of the plates, A = πr² = 1.5 m²
Thus, the maximum charge that can be placed on disks without a spark is:
Q = E×ε₀×A
⇒ Q = 3 × 10⁶ N/C × 8.85 × 10⁻¹² F/m × 1.5 m² = 39.8 × 10⁻⁶ C = 39.8 μC.