your answer is make up artist
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
Answer:
weightlessness, condition experienced while in free-fall, in which the effect of gravity is canceled by the inertial (e.g., centrifugal) force resulting from orbital flight. ... Excluding spaceflight, true weightlessness can be experienced only briefly, as in an airplane following a ballistic (i.e., parabolic) path.
Answer:
v = 2.18m/s
Explanation:
In order to calculate the speed of Betty and her dog you take into account the law of momentum conservation. The total momentum before Betty catches her dog must be equal to the total momentum after.
Then you have:
(1)
M: mass Betty = 40kg
m: mass of the dog = 15kg
v1o: initial speed of Betty = 3.0m/s
v2o: initial speed of the dog = 0 m/s
v: speed of both Betty and her dog = ?
You solve the equation (1) for v:
The speed fo both Betty and her dog is 2.18m/s
