Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled
Let
α = the angular acceleration
ω = the final angular velocity
The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²
The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s
If the thrower's arm is r meters long, the tangential velocity of release will be
v = 13.963r m/s
Answer: 13.963 rad/s
Using the formula v=f times lambada
then v=the speed of light.
and f=what’s we’re looking for
and lambada=the wavelength.
so then you sub what you have (v and lambada) in the formula.
then multiply the frequency(f) by the given wavelength and then solve for f
The object's speed will not change.
In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:

where the first term is the resultant of the forces acting on the body, m is the mass of the object and a its acceleration, we see that if no forces act on the object, then the acceleration is zero. Therefore, the acceleration of the object is zero, and its velocity remains constant.
Answer:
Mass of bullet is m=0.01kg
Mass of the block is M=4kg
Coefficient=0.25,distance=20m
So, let the speed of the block just after the bullet embedded in it be V and v be the speed of bullet before striking the block,
By applying conservation of momentum,
mv=(m+M)V
V=
M+m
mv
Explanation:
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Answer:
given , v = 300 km/hr; distance d = 1500 km; then time t = d/v = 1500/300 = 5 hrs
Explanation: