The final velocity of the truck is found as 146.969 m/s.
Explanation:
As it is stated that the lorry was in standstill position before travelling a distance or covering a distance of 3600 m, the initial velocity is considered as zero. Then, it is stated that the lorry travels with constant acceleration. So we can use the equations of motion to determine the final velocity of the lorry when it reaches 3600 m distance.
Thus, a initial velocity (u) = 0, acceleration a = 3 m/s² and the displacement s is 3600 m. The third equation of motion should be used to determine the final velocity as below.
Then, the final velocity will be
Thus, the final velocity of the truck is found as 146.969 m/s.
Answer:
(a) 41.75m/s
(b) 4.26s
Explanation:
Let:
Distance, D = 89m
Gravity, = 9.8 m/
Initial Velocity, = 0m/s
Final Velocity, = ?
Time Taken, = ?
With the distance formula, which is
D = +
and by substituting what we already know, we have:
89 = ×9.8×
With the equation above, we can solve for :
Now that we have solved , we can use the following velocity formula to solve for :
, where is also equals to , so we have
By substituting , , and ,
We have:
Answer:
Incomplete question, check attachment for the graph needed to solve problem.
A 8.1nm........
Explanation:
Electric Field is given as
E=V/d
Where V is voltage
And d is the distance apart
E is the electric field
The voltage V just before action of potential is -70mV,
The value d=8.1nm
d=8.1×10^-9m
E=V/d
E=-70×10^-3/8.1×10^-9
E=-8.6×10^6 N/C
Then the magnitude of the electric field is 8.6×10^6N/C