Answer:
Force exerted on the car is 7030 N.
Explanation:
It is given that,
Mass of the car, m = 740 kg
Initial speed of the car, u = 19 m/s
Final speed of the car, v = 0
Time taken, t = 2 s
Let F is the force exerted on the car during this stop. We know that it is equal to the product of force and acceleration. Mathematically, it is given as :
F=m\times \dfrac{v-u}{t}F=m×tv−u
F=740\times \dfrac{0-19}{2}F=740×20−19
F = -7030 N
So, the force exerted on the car during this stop is 7030 N. Hence, this is the required solution.
Answer:
Hi there!
There are many possible answers!
My best guesses are:
1) Knowing how they work will prevent injury! For example, knowing that the gears twist will stop you from putting your finger in it!
2) Allowing you to use it! Knowing how to line up the edge of the can with the tool will let you use it properly!
Hope this helps
Work done = mgh where h varies from 0 (at the bottom of the pool) to 5 ft (at the sides of the pool).
m= density*volume = 62.5*pi*24^2*d/4 = 9000πd lbs
h = 5-d (the particles at the top of the pool require less work as required to those at the bottom).
Then,
WD (E) = 9000πd*9.81*(5-d) = 88290πd(5-d) = 441450πd-88290πd^2
Integrating from 0-4 with respect to depth of pool, d
Total E = {441450π*d^2/2 - 88290π*d^3/} for d between 0 and 4 ft
E=220725π*d^2 -29430π*d^3
Substituting for d=0 ft and d=4 ft;
= {220725*π*4^2-29430π*4^3} - {220725π*0^2-29430*pi*2*0^3)
=5177596.021 J= 5.17 MJ