Answer:
1.8m
Explanation:
Let the Elastics of the steel ASTM-36 
The strain of the bar when subjected to 150 MPa is
Therefore, if the bar elongates by 1.35 mm, then the original length L would be:
or 1.8m
it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
Answer: 
Explanation:
Given
Current in the first wire 
Current in the second wire 
wires are 
apart
Force per unit length between the current-carrying wires is
Force exerted by the wires is the same
Put the values
This force will be repulsive in nature as the current is flowing opposite
Answer:
160000000 kg.
Explanation:
p=mv
p=1.6x10^9
v=10m/s
rearrange and substitute:
(1.6x10^9)=m(10)
m=(1.6x10^9)/10
m= 1.6x10^8 kg.
