You can compare the velocity of the car, 60 mph, with the velocity that a mass would acquire when falls from certain height.
First, convert 60 mph to m/s:
60 miles/h * 1.60 km/mile * 1000 m/km * 1h/3600s = 26.67 m/s
Second, calculate from what height a body in free fall reachs 26.67 m/s velocity when hits the floor.
free fall => Vf^2 = 2g*H => H = Vf^2 / (2g)
H = (26.67m/s)^2 / (2*9.8 m/s) = 36.2 m
If you consider that the height between the floors of a building is approximately 3.6 m, you get 36.2 m / 3.6 m/floor = 10 floors.
Then, you conclude that the force of impact is the same as driving you vehicle off a 10 story building.
1. Define Newtons second law of motion (this will help put things into perspective)
2.Get the mass of the object (in this case 75 kg)
3.The net force acting on the object...find it (in this case 500 N)
4.Change the equation to F=ma (500=75a)
5.Divide both sides by 75 and that is the acceleration.
Jumping on a trampoline is a classic example of conservation of energy, from potential into kinetic. It also shows Hooke's laws and the spring constant. Furthermore, it verifies and illustrates each of Newton's three laws of motion.
<u>Explanation</u>
When we jump on a trampoline, our body has kinetic energy that changes over time. Our kinetic energy is greatest, just before we hit the trampoline on the way down and when you leave the trampoline surface on the way up. Our kinetic energy is 0 when you reach the height of your jump and begin to descend and when are on the trampoline, about to propel upwards.
Potential energy changes along with kinetic energy. At any time, your total energy is equal to your potential energy plus your kinetic energy. As we go up, the kinetic energy converts into potential energy.
Hooke's law is another form of potential energy. Just as the trampoline is about to propel us up, your kinetic energy is 0 but your potential energy is maximized, even though we are at a minimum height. This is because our potential energy is related to the spring constant and Hooke's Law.
A plane mirror always forms a virtual image. the image and the object are the same distance from a flat mirror, the image size is the same as the object, and the image is upright!
It is given that for the convex lens,
Case 1.
u=−40cm
f=+15cm
Using lens formula
v
1
−
u
1
=
f
1
v
1
−
40
1
=
15
1
v
1
=
15
1
−
40
1
v=+24.3cm
The image in formed in this case at a distance of 24.3cm in left of lens.
Case 2.
A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that
u=∞
f=15cm
Now, using mirror’s formula
v
1
+
u
1
=
f
1
v
1
+
∞
1
=
15
1
v=+15cm
The image is formed at a distance of 15cm in left of mirror
