Of the cliff?
Projectile motion
In the problem we are asked to find a height of certain cliff when a motorcycle stunt driver zoom out horizontally at the end the cliff at an initial velocity. So we will use one of the kinematics equation for projectile motion,
y
=
v
o
y
t
+
1
2
g
t
where
v
o
y
is just equal to zero since we can assume that the driver zooms out horizontally,
g
=
9.8
m
/
s
2
and
t
is time after
That would be reflection.
Answer:
The maximum speed will be 26.475 m/sec
Explanation:
We have given mass of the toy m = 0.50 kg
radius of the light string r = 1 m
Tension on the string T = 350 N
We have to find the maximum speed without breaking the string
For without breaking the string tension must be equal to the centripetal force
So 
So 

v = 26.475 m /sec
So the maximum speed will be 26.475 m/sec
Answer:
mass consumed by 235U each day = 2 kg
Explanation:
electrical power produced = 1 GW = 1 × 10⁹ × (6.24151 × 10¹⁸ ) eV
= 6.24151× 10²¹ MeV/s
thermal energy = 0.420 * 250 = 105 MeV

= 5.94 × 10¹⁹ fission/second
=5.94 × 10¹⁹× 24 × 60 ×60)
= 5.13 × 10²⁴ fission/day
mu = 235.04393 × 1.660× 10 ⁻²⁷ = 390.1729× 10⁻²⁷ Kg
M = mu ×5.13 × 10²⁴
= 390.1729× 10⁻²⁷ ×5.13 × 10²⁴
M = 2 kg(approx.)
mass consumed by 235U each day = 2 kg