<em>IF</em> this is happening on <u>Earth</u>, then the acceleration of gravity wherever the captain and the roof are is about 9.8 m/s² downward.
<em>IF</em> his shield is not affected by air resistance, and it's free to behave under the influence of gravity alone, then its acceleration is <em>9.8 m/s² downward</em>, from the moment it leaves his hand until the moment it hits a building, a tree, a car, a bus, an evildoer, or the ground.
Whatever speed it has when it leaves his hand, in whatever direction, makes no difference.
Answer:
The magnitude of the average angular acceleration is calculated as
Explanation:
Maximum speed that can be attained by the disk, = 10,000 rpm
Speed of spinning of the disk, N = 7570 rpm
Time taken to come to rest, t = 0.435 s
Now,
The initial angular velocity is given by:
Final angular velocity,
The average angular acceleration of the disk can be computed by using the kinematic eqn:
Answer:
B. 0.98 m/s
Explanation:
This is because we use the simple formula of dividing the distance by the time. In which case would be 13.69m (distance) divided by 13.92s (time) and we will get 0.983477011 or 0.98m/s (your answer)
I hope this made sense and hoped it helped. Good luck with your test luv :)
Answer:
speed of the bullet before it hit the block is 200 m/s
Explanation:
given data
mass of block m1 = 1.2 kg
mass of bullet m2 = 50 gram = 0.05 kg
combine speed V= 8.0 m/s
to find out
speed of the bullet before it hit the block
solution
we will apply here conservation of momentum that is
m1 × v1 + m2 × v2 = M × V .............1
here m1 is mass of block and m2 is mass of bullet and v1 is initial speed of block i.e 0 and v2 is initial speed of bullet and M is combine mass of block and bullet and V is combine speed of block and bullet
put all value in equation 1
m1 × v1 + m2 × v2 = M × V
1.2 × 0 + 0.05 × v2 = ( 1.2 + 0.05 ) × 8
solve it we get
v2 = 200 m/s
so speed of the bullet before it hit the block is 200 m/s
C
Because in my opinion they do bending of a light wave as it passes at an angle from one medium to another.