The energy stored in a capacitor is given by:
where
U is the energy
C is the capacitance
V is the potential difference
The capacitor in this problem has capacitance
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
Answer:
2 kg
Explanation:
Acceleration = 5 m/s^2
Force = 10 N
Force = mass * acceleration
mass = force / acceleration
mass = 10 / 5
mass = 2 kg
As we know that the formula of kinetic energy will be
now here we know that
m = 2 kg
v = 1 m/s
so from the above equation we have
D = distance between the cars at the start of time = 680 km
v₁ = speed of one car
v₂ = speed of other car = v₁ - 10
t = time taken to meet = 4 h
distance traveled by one car in time "t" + distance traveled by other car in time "t" = D
v₁ t + v₂ t = D
(v₁ + v₂) t = D
inserting the values
(v₁ + v₁ - 10) (4) = 680
v₁ = 90 km/h
rate of slower car is given as
v₂ = v₁ - 10
v₂ = 90 - 10 = 80 km/h
Answer:
B
Explanation:
From Newton's law of motion, we have:
V^2 = U^2 + 2gH
Where V and U are final and initial velocity respectively.
H is the height.
For the object to have a sustain a maximum height it means the final velocity of the object is zero.
By computing the height of the object sustain by A, we have:
0^2 = 2^2 -2×10×H
0= 4 -20H
4 = 20H;
H= 0.2m
For object B we have;
0^2 = 1^2 -2×10×H
0 = 1 -20H
H = 1/20= 0.05m
From computing the height sustain by both objects, we see object B is projected at a shorter height into atmosphere than A.
Hence object B will return to the ground first.
