Answer:
The final rotation is 2.449 times the initial rotation
Explanation:
Centripetal acceleration and rotation speed of particle is related as
a=rω²
where ω is angular speed and r radius of circle
Rearranging equation
ω=
The initial acceleration is given by
a₁ =100g
The final acceleration of particle is
a₂ = 6 (100g)
=600g
The final and initial rotation speed are related as

ω2=
ω2=2.449ω1
Answer:
10.4 m/s
Explanation:
The problem can be solved by using the following SUVAT equation:

where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
For the diver in the problem, we have:
is the initial velocity (positive because it is upward)
is the acceleration of gravity (negative because it is downward)
By substituting t = 1.7 s, we find the velocity when the diver reaches the water:

And the negative sign means that the direction is downward: so, the speed is 10.4 m/s.
Answer:
v= 2413.5 m/s
Explanation:
maximum change of speed of rocket
=(initial exhaust velocity)×ln [(initialmass/finalmass)]
let initial mass= m
final mass = m-m(4/5) = m/5
[since the 80% of mass which is fuel is exhausted]
V-0 = 1500 ln (1/0.2)
V= 1500×1.609 = 2413.5 m/s
therefore, its exhaust speed v= 2413.5 m/s
Explanation:
Given that,
Force with which a child hits a ball is 350 N
Time of contact is 0.12 s
We need to find the impulse received by the ball. The impulse delivered is given by :

So, the impulse is 42 N-m..
We know that he change in momentum is also equal to the impulse delivered.
So, impulse = 42 N-m and change in momentum =42 N-m.
Answer:
29.4 N/m
0.1
Explanation:
a) From the restoring Force we know that :
F_r = —k*x
the gravitational force :
F_g=mg
Where:
F_r is the restoring force .
F_g is the gravitational force
g is the acceleration of gravity
k is the constant force
xi , x2 are the displacement made by the two masses.
Givens:
<em>m1 = 1.29 kg</em>
<em>m2 = 0.3 kg </em>
<em>x1 = -0.75 m </em>
<em>x2 = -0.2 m </em>
<em>g = 9.8 m/s^2 </em>
Plugging known information to get :
F_r =F_g
-k*x1 + k*x2=m1*g-m2*g
k=29.4 N/m
b) To get the unloaded length 1:
l=x1-(F_1/k)
Givens:
m1 = 1.95kg , x1 = —0.75m
Plugging known infromation to get :
l= x1 — (F_1/k)
= 0.1