Answer:
540C.
Explanation:
A capacitor of capacitance C when charged to a voltage of V will have a charge Q given as follows;
Q = CV ----------(i)
From the question, the initial charge on the capacitor is the charge on it before it was connected to the resistor. In other words, the initial charge on the capacitor will have a maximum value which can be calculated using equation (i) above.
Where;
C = 6F
V = 90V
Substitute these values into equation (i) as follows;
Q = 6 x 90
Q = 540 C
Therefore, the initial charge on the capacitor is 540C.
If the object is in equilibrium that means that the sum of the forces on it is zero and the net force is zero. If none of the forces changes then the object continues in constant uniform motion. That means constant speed in a straight line.
Answer:
4.44 i + 11.11 j , 11.96 m/s
Explanation:
m1 = 0.2 kg, v1 = 10 m/s along X axis, v1 = 10 i
m2 = 0.25 kg, v2 = 20 m/s along north, v2 = 20 j
After collision they stick together and let they move with velocity v.
Use the conservation of momentum
m1 x v1 + m2 x v2 = (m1 + m2) x v
0.2 x (10 i) + 0.25 x (20 j) = (0.2 + 0.25) x v
2i + 5j = 0.45 v
v = 4.44 i + 11.11 j
magnitude of velocity = = 11.96 m/s
Thus, the velocity of 0.250 kg hockey puck is 4.44 i + 11.11 j metre per second and the magnitude of velocity is 11.96 m/s.
The subway train is moving to the left and then comes to a stop, what direction is the acceleration will be towards right and will be positive.
The train comes to a stop, this means that the velocity gradually decreases. Hence, the acceleration must be positive and points towards the right.
velocity and acceleration:
velocity is the charge of alternate inside the position of an object, at the same time as acceleration is the rate of trade of the fee. consider the movement alongside a line. If the velocity and acceleration are in the same route, then the charge constantly will growth in time. inside the meantime, if the speed and acceleration are in contrary instructions, then the rate steadily decreases to 0 speed, reverses route, and now the velocity is in the identical path of the acceleration.
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