Answer:
Explanation:
The work W required for the transfer is equal to the energy difference
where q is the charge we want to transfer and ΔV is the potential difference.
So, the work needed will be
Now, the average power is:
As the Watt is Joule/second, we need the 45 min multiplied by 60:
Taking all this together:
or
Answer:
285 Litre ( 1kg of water occupies 1 Litre )
which is identical to 285 dm^3 ie 285 cubes of length 10 cm
and as there are 10 of these linearly in 1 m then there are 10^3 = 1000 dm^3 in each m^3
giving a volume of 0.285 m^3
Of course if you KNEW that 1m^3 of water has a mass of 1000 kg then you would simply have said that the volume was 285/1000 m^3
The problem about describes a perfectly inelastic collision. We are tasked to find the initial velocity of an object having a mass of 6 kg moving due west. It is given in the problem that after collision the cart sticks together and it stops. Thus, the final mass is the sum of the two cart and the final velocity is zero. For a perfectly inelastic collision,
m1v1-m2v2=vf(m1+m2)
By Substitution,
3(4)-6(v2)=0
6v2=12
v2=2
Therefor, the initial velocity if a 6 kg cart is 2 m/s
Answer:
u= 35.30 s
Explanation:
given,
horizontal distance covered by the ball = 105 m
vertical distance to clear by the ball = 4 m
angle at which the ball was hit = 31°
Let the initial velocity is u m/s and the ball take t sec to reach the fence.
ut = 122.5
Using
t = 3.47 s
now,
u= 35.30 s
Speed of the ball when it leaves the bat u= 35.30 s
Answer:
1) T = 4.5 s
2) T = 4.5 s
3) v = 9.9 m/s
Explanation:
We can use the equation
T = 2π√(L/g)
1) T = 2π√(5m/9.81 m/s²) = 4.5 s
2) T = 2π√(L/g)
T = 2π√(5m/9.81 m/s²) = 4.5 s
3) v = √(2gR)
v = √(2(9.81 m/s²)(5m))
v = 9.9 m/s