Answer:
inelastic, since the girl moves in the same direction as the thrown ball
Explanation:
yess this ok
UwU
Explanation:
Show that the motion of a mass attached to the end of a spring is SHM
Consider a mass "m" attached to the end of an elastic spring. The other end of the spring is fixed
at the a firm support as shown in figure "a". The whole system is placed on a smooth horizontal surface.
If we displace the mass 'm' from its mean position 'O' to point "a" by applying an external force, it is displaced by '+x' to its right, there will be elastic restring force on the mass equal to F in the left side which is applied by the spring.
According to "Hook's Law
F = - Kx ---- (1)
Negative sign indicates that the elastic restoring force is opposite to the displacement.
Where K= Spring Constant
If we release mass 'm' at point 'a', it moves forward to ' O'. At point ' O' it will not stop but moves forward towards point "b" due to inertia and covers the same displacement -x. At point 'b' once again elastic restoring force 'F' acts upon it but now in the right side. In this way it continues its motion
from a to b and then b to a.
According to Newton's 2nd law of motion, force 'F' produces acceleration 'a' in the body which is given by
F = ma ---- (2)
Comparing equation (1) & (2)
ma = -kx
Here k/m is constant term, therefore ,
a = - (Constant)x
or
a a -x
This relation indicates that the acceleration of body attached to the end elastic spring is directly proportional to its displacement. Therefore its motion is Simple Harmonic Motion.
Answer:
a)Velocity of car =v=16 m/s
b)Force against the track at point B=1.15*
N
Explanation:
Given mass of roller coaster=m=350 kg
Position of A=Ha=25 m
Position of B=Hb=12 m
Net potential energy=mg(ha-hb)
Net potential energy=(350)(9.80)(25-12)
Net potential energy=44590 J
Using energy conservation
net kinetic energy=net potential energy
(1/2)mv^2=mg(ha-hb)
m=350
velocity=v=16 m/s
b)There two force acting,centripetal force upward and gravity downward.
Thus net force acting will be
Net force=(mv^2/r)-mg
Net force=14933.33-3430
Net force=1.15* N
Answer:
102000 kg
Explanation:
Given:
A total Δν = 15 km/s
first stage mass = 1000 tonnes
specific impulse of liquid rocket = 300 s
Mass flow rate of liquid fuel = 1500 kg/s
specific impulse of solid fuel = 250 s
Mass flow of solid fuel = 200 kg/s
First stage burn time = 1 minute = 1 × 60 seconds = 60 seconds
Now,
Mass flow of liquid fuel in 1 minute = Mass flow rate × Burn time
or
Mass flow of liquid fuel in 1 minute = 1500 × 60 = 90000 kg
Also,
Mass flow of solid fuel in 1 minute = Mass flow rate × Burn time
or
Mass flow of solid fuel in 1 minute = 200 × 60 = 12000 kg
Therefore,
The total jettisoned mass flow of the fuel in first stage
= 90000 kg + 12000 kg
= 102000 kg
