Answer:
V at C is 3.6 m/s
Explanation:
At A kinetic energy is zero and potential energy=mgh=0.5*9.81*0.6=2.943 J
By conservation of energy.
KE+PE=Constant
At C PE=0.6 J
the KE=2.943-0.6=2.343 J
KE=0.5*m*v^2
v=√[KE/(0.5*m)]=3.06 m/s
Answer:
Power=V*I which corresponds to the second option shown: "voltage times amperage"
Explanation:
The electric power is the work done to move a charge Q across a given difference of potential V per unit of time.
Since such electrical work is the product of the potential difference V times the charge that moves through that potential, and this work is to be calculated by the unit of time, we need to divide the product by time (t) which leads to the following final simple equation

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The maximum velocity in a banked road, ignoring friction, is given by;
v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.
Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°
Therefore, the road has been banked at 5.24°.