Answer:
Distance of the point where electric filed is 2.45 N/C is 1.06 m
Explanation:
We have given charge per unit length, that is liner charge density 
Electric field E = 2.45 N/C
We have to find the distance at which electric field is 2.45 N/C
We know that electric field due to linear charge is equal to
, here 
is linear charge density and r is distance of the point where we have to find the electric field
So 
r = 1.06 m
So distance of the point where electric filed is 2.45 N/C is 1.06 m
The rate of acceleration of the crate would be 1 m/s^2 because the equation for force is F=ma and when you plug in your numbers you get 10=10a so a=1
Answer:
The initial velocity was U=22.14m/s
Explanation:
Step one :
Applying the third equation of motion
v² = u²+ 2as
Where v= Final velocity
U =initial velocity
a= acceleration due to gravity
S= distance or displacement
Step two :
V= 0
a= 9.81m/s²
S=25m
U=?
Step three :
Substituting into the equation we have
0²=U²+2*9.81*25
0=U²+490.5
U²=-490.5
U=√490.5
U=22.14m/s
At the bottom of the rotation, the kinetic far exceeds the potential. However, at both tops, potential exceeds kinetic.
