A 1,800 kg car is parked on a road that has an elevation level of 7°. Suppose the coefficient of static friction is .65. What is
the force of static friction?
1 answer:
m = mass of the car = 1800 kg
g = acceleration due to gravity = 9.8 m/s²
= force of gravity on the car
force of gravity on the car is given as
= mg
inserting the values
= (1800) (9.8) = 17640 N
= static frictional force acting on the car
from the force diagram of the car , the static frictional force is opposite to the parallel component of force of gravity on the car . hence static frictional force mus balance the component of force of gravity parallel to incline surface.
hence
=
Sin7
= (17640) Sin7
= 2149.8 N
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Answer:
2.2 meters
Explanation:
Potential energy, PE created by a charge, q at a radius r from the charge source, Q, is expressed as:
is Coulomb's constant.
#The electric field, at radius r is expressed as:
From i and ii, we have:
#Substitute actual values in our equation:
Hence, the distance between the charge and the source of the electric field is 2.2 meters