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
Answer:11 km/s
Explanation:
Given
Escape velocity at the surface of earth is 11 km/s
Escape velocity is given by

Escape velocity at the surface of earth
--------------------1
If Escape velocity is three times and the radius is also the three times


i.e. 
Answer:

Explanation:
The mass of the man can be found by using the formula

f is the force
a is the acceleration
From the question we have

We have the final answer as
<h3>50 kg</h3>
Hope this helps you
Explanation:
Given data:
d = 30 mm = 0.03 m
L = 1m
S
= 70 Mpa
Δd = -0.0001d
Axial force = ?
validity of elastic deformation assumption.
Solution:
O'₂ = Δd/d = (-0.0001d)/d = -0.0001
For copper,
v = 0.326 E = 119×10³ Mpa
O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶
∵δ = F.L/E.A and σ = F/A so,
σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa
F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN
S
= 70 MPa > σ = 36.5 MPa
∵ elastic deformation assumption is valid.
so the answer is
F = 25800 K N and S
> σ