The acceleration of the car will be needed in order to calculate the time. It is important to consider that the final speed is equal to zero:

We can clear time in the speed equation:

If you find some mistake in my English, please tell me know.
Answer:
C. 0.2 Hertz
Explanation:
The frequency of a spring is equal to the reciprocal of the period:

where
f is the frequency
T is the period
For the spring in this problem,
T = 5 s
therefore, the frequency is

Answer:
Electric field, E = 0.064 V/m
Explanation:
It is given that,
Resistivity of silver wire, 
Current density of the wire, 
We need to find the magnitude of the electric field inside the wire. The relationship between electric field and the current density is given by :


E = 0.0636 V/m
or
E = 0.064 V/m
So, the magnitude of electric field inside the wire is 0.064 V/m. Hence, this is the required solution.
Answer:
The direction of electric field and equipotential line at the same point are always PERPENDICULAR TO THE ELECTRIC FIELD.
Explanation:
Equipotential surface is a three dimensional part of equipotential lines.
Equipotential lines are a type of contour lines that is use to trace lines that have the same altitude on the map and the altitude is the electric potential.
Equipotential lines are always perpendicular to electric potential because the lines creates three dimension equipotential surface.
Answer:
scientific law is a statement that summarizes a pattern found in nature.