Answer:
Explanation:
It is given that,
Speed of one quoll around a curve, v = 3.2 m/s (maximum speed)
Radius of the curve, r = 1.4 m
On the curve, the centripetal force is balanced by the frictional force such that the coefficient of frictional is given by :
So, the coefficient of static friction between the quoll's feet and the ground in this trial is 0.74. Hence, this is the required solution.
Answer:
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Answer:
<em>The final charge on the 6.0 mF capacitor would be 12 mC</em>
Explanation:
The initial charge on 4 mF capacitor = 4 mf x 50 V = 200 mC
The initial Charge on 6 mF capacitor = 6 mf x 30 V =180 mC
Since the negative ends are joined together the total charge on both capacity would be;
q =
q = 200 - 180
q = 20 mC
In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage
q = (4 x V) + (6 x V)
20 = 10 V
V = 2 V
For the final charge on 6.0 mF;
q = CV
q = 6.0 mF x 2 V
q = 12 mC
Therefore the final charge on the 6.0 mF capacitor would be 12 mC