The car has gravitational potential energy (Eg)
Eg = mgh
Eg = (53kg) (9.81N/kg) (14m)
Eg = 7 279.02 J
1) First, let's calculate the value of deceleration a that the car can achieve, using the following relationship:

where S=67 m is the distance covered, vf=0 is the final velocity of the car, and vi=15 m/s is the initial velocity. From this we can find a:

2) Then, we can assume this is the value of acceleration that the car is able to reach. In fact, the force the brakes are able to apply is

This force will be constant, and since m is always the same, then a is the same even in the second situation.
3) Therefore, in the second situation we have a=-1.68 m/s^2. However, the initial velocity is different: vi=45 m/s. Using the same formula of point 1), we can calculate the distance covered by the car before stopping:

Answer:
weigh is 2353.13 N
Explanation:
Given data
density = 7890.00 kg/m3
lighter = 299 N
to find out
the volume of the anchor and weigh in air
solution
from question we can say that
apparent weight = actual weight - buoyant force
we know weight = mg and buoyant force = water density × g
so volume of anchor is = actual weight - apparent weight / buoyant force
volume of anchor is = 299 / 1000 × 9.81
volume of anchor is = 0.0304791 m³
and
weight of anchor is mg
here mass m = density Fe g
density Fe = 7870 from table 14-1
so weight = 7870 × 0.0304791 × 9.81
weigh is 2353.13 N