Answer:
The compression in the spring is 5.88 meters.
Explanation:
Given that,
Mass of the car, m = 39000 kg
Height of the car, h = 19 m
Spring constant of the spring, 
We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,
x is the compression in spring
So, the compression in the spring is 5.88 meters.
Answer:
They will meet at a distance of 7.57 m
Given:
Initial velocity of policeman in the x- direction, 
The distance between the buildings, 
The building is lower by a height, h = 2.5 m
Solution:
Now,
When the policeman jumps from a height of 2.5 m, then his initial velocity, u was 0.
Thus
From the second eqn of motion, we can write:
t = 0.707 s
Now,
When the policeman was chasing across:
The distance they will meet at:
9.57 - 2.0 = 7.57 m
Use conservation of momentum ;
m1u1 + m2u2 = m1v1 + m2v2
1200×15.6 + 0 = 2700v
v = 18720/2700
v = 6.933 or ~ 7 m/s
Answer:
B - Divide Speed and Distance
Explanation:
Hope it helps!!
:D
Answer:
Y = 78.13 x 10⁹ Pa = 78.13 GPa
Explanation:
First we will find the centripetal force acting on the wire as follows:
F = mv²/r
where,
F = Force = ?
m = mass of rock = 0.34 kg
v = speed = 19 m/s
r = length of wire
Therefore,
F = (0.34)(19)²/r
F = 122.74/r
now, we find cross-sectional area of wire:
A = πd²/4
where,
A = Area = ?
d = diameter of wire = 1 mm = 0.001 m
Therefore,
A = π(0.001)²/4
A = 7.85 x 10⁻⁷ m²
Now, we calculate the stress on wire:
Stress = F/A
Stress = (122.74/r)/(7.85 x 10⁻⁷)
Stress = 1.56 x 10⁸/r
Now, we calculate strain:
Strain = Δr/r
where,
Δr = stretch in length = 2 mm = 0.002 m
Therefore,
Strain = 0.002/r
now, for Young's modulus (Y):
Y = Stress/Strain
Y = (1.56 x 10⁸/r)/(0.002/r)
<u>Y = 78.13 x 10⁹ Pa = 78.13 GPa</u>
