Answer:
75 m
Explanation:
The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.
The horizontal component of the velocity of the projectile is

and it is constant during the motion;
the total time of flight is
t = 5 s
Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

Answer:
The final velocity of the bullet is 9 m/s.
Explanation:
We have,
Mass of a bullet is, m = 0.05 kg
Mass of wooden block is, M = 5 kg
Initial speed of bullet, v = 909 m/s
The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,

So, the final velocity of the bullet is 9 m/s.
Answer:
-.457 m/s^2
Explanation:
Actual weight = 60 .3 (9.81) = 591.54 N
Accel of lift changes this to 60.3 ( 9.81 - L) where L - accel of lift
60.3 ( 9.81 - L ) = 564
solve for L = .457 m/s^2 DOWNWARD
so L = - .457 m/s^2
Answer:
unmmmmmmmm I think the answerA
The first one is dependent variable
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