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
<span>Without friction, there will be undamped simple harmonic motion. The force of the spring is proportional to the distance from the equilibrium point. The period of oscillation will be independent of the amplitude.
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Answer:
I don't know
Explanation:
you need then search here
Answer:
14,700 N
Explanation:
The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:
(1)
the weight of the hyppo is

where m is the hyppo's mass and g is the gravitational acceleration; therefore, solving eq.(1) for N, we find

Answer:
linear charge density = -9.495 ×
C/m
Explanation:
given data
revolutions per second = 1.80 ×
radius = 1.20 cm
solution
we know that when proton to revolve around charge wire then centripetal force is require to be in orbit of radius around provide by electric force
so
- q × E = m × w² × r ..................1
- 9 ×
×
q = m × w² × r ............2
and w =
w =
w = 1.80 ×
×
w = 11304000 rad/s
so here from equation 2
- 9 ×
×
1.80 ×
= 1.672 ×
× 11304000² × 0.0120
linear charge density = -9.495 ×
C/m