Answer:
8.505 m
Explanation:
Let V1 and V2 be velocities of puck A and B respectively
Since A and B move in the same direction, so the relative velocity will be V1+V2=3.5+3.9=7.4m/s
Or
Vr=7.4 m/s
Distance=S= 18 m
Time =t=?
S=Vr×t
==> t=S/Vr
==> t= 18/7.4=2.43 sec
At this time both will strike together
<em><u>Distance by puck A</u></em>
<em>V1=3.5 m/s</em>
Time=t= 2.43 sec
Distance covered=d=?
d=V1×t=3.5×2.43=8.505 m
So, puck A will cover 8.505 meters before collision
Answer:
True.
Explanation:
Newton's First Law of Motion states that every object continues in it's state of rest or of uniform motion in a straight line unless acted upon by an external force.
The answers to all three questions depend on WHICH branch
of Physics you're talking about . . . a piece of information you've
neglected to specify.
For this case we first think that the skateboard and the child are one body.
We have then:
1 = jug
2 = skateboard + boy
By conservation of the linear amount of movement:
M1V1i + M2V2i = M1V1f + M2V2f
Initial rest:
v1i = v2i = 0
0 = M1V1f + M2V2f
Substituting values
0 = (7.8) (3.2) + (M2) (- 0.65)
0 = 24.96 + M2 (-0.65)
-24.96 = (-0.65) M2
M2 = (-24.96) / (- 0.65) = 38.4 kg
Then, the child's mass is:
M2 = Mskateboard + Mb
Clearing:
Mb = M2-Mskateboard
Mb = 38.4 - 1.9
Mb = 36.5 Kg
answer:
the boy's mass is 36.5 Kg
The other two bulbs stay lit with the same brightness.