Answer:
c. 0.02 C and 4 J
Explanation:
Applying,
Q = CV................ Equation 1
Where Q = Charge, C = Capacitance of the capacitor, V = Voltage.
From the question,
Given: C = 50 μF = 50×10⁻⁶ F, V = 400 V
Substitute these values into equation 1
Q = (50×10⁻⁶)(400)
Q = 0.02 C.
Also Applying
E = CV²/2............. Equation 2
Where E = Energy stored.
Therefore,
E = (50×10⁻⁶ )(400²)/2
E = 4 J
Hence the right option is c. 0.02 C and 4 J
Answer:
the required minimum magnitude of the force F is 21 N
Explanation:
Given the data in the question,
m = 5 kg
width = 60 cm
height = 80 cm
Let force is F represent in the image below,
so when the block about to rotate normal shifted to edge of cube
mg(w/2) = Fh
F = mg(w/2) / h
we know that g = 9.8 m/s²
we substitute
F = (5 × 9.8 ( 60/2)) / 70
F = (5 × 9.8 × 30 ) / 70
F = 1470 / 70
F = 21 N
Therefore, the required minimum magnitude of the force F is 21 N
Answer:
the vibrations push the purse up and down very fast and gravity pushes the purse down onto the floor
Explanation: does that help
its B 0.225kPa using the formula p=f/A then change the pascals into kpa