To look for the acceleration, it will come from:
vf^2=v0^2+2ad
where:
vf = final velocity = 0
v0 = initial velocity =251 m/s
a = acceleration
d= distance traveled = 0.237 m
0=251^2+2a(0.237 )
a= -251 ^2 / (2*0.237) =-132 913.502 m/s/s
we find the force from:
F = ma = 0.0115kg*(-1.32x10^5m/s/s) = -1518 N
the negative sign shows that the force is in the direction contradictory the
bullet's motion
Answer:
If the ray of light is deflected by 45 degrees by the first mirror its total deflection by mirror (I) is 90 deg. (incident = 45 and exit ray equals 45 deg)
The second mirror will cause a net deflection of 90 degrees and the total deflection will be 180 deg or in opposite direction to the incident ray.
Answer:
The work done by the drag force is given by 29.96 J
Explanation:
Given :
Thrust force
N
Displacement
m
Mass of rocket
Kg
From work energy theorem,


Where
thrust work
gravitational work

After cutoff kinetic energy is converted into potential energy,

Put value of KE

Work done by drag force is given by,

J
Therefore, the work done by the drag force is given by 29.96 J