The net force acting on the car is equal to -12000 Newton.
<u>Given the following data:</u>
- Initial velocity = 30 m/s.
- Final velocity = 0 m/s (since the car came to a stop).
To determine the net force acting on the car:
<h3>How to calculate the net force.</h3>
Mathematically, the net force acting on the car is given by Newton's Second Law of Motion:
<u>Where:</u>
- u is the initial velocity.
Substituting the given parameters into the formula, we have;
Net force = -12000 Newton.
Read more on net force here: brainly.com/question/1121817
The other of the four organic molecules is carbohydrates.
All four of them are:
Nucleic Acids (DNA and RNA stuff)
Proteins
Lipids
& Carbohydrates
Let's cut through the weeds and the trash
and get down to the real situation:
A stone is tossed straight up at 5.89 m/s .
Ignore air resistance.
Gravity slows down the speed of any rising object by 9.8 m/s every second.
So the stone (aka Billy-Bob-Joe) continues to rise for
(5.89 m/s / 9.8 m/s²) = 0.6 seconds.
At that timer, he has run out of upward gas. He is at the top
of his rise, he stops rising, and begins to fall.
His average speed on the way up is (1/2) (5.89 + 0) = 2.945 m/s .
Moving for 0.6 seconds at an average speed of 2.945 m/s,
he topped out at
(2.945 m/s) (0.6 s) = 1.767 meters above the trampoline.
With no other forces other than gravity acting on him, it takes him
the same time to come down from the peak as it took to rise to it.
(0.6 sec up) + (0.6 sec down) = 1.2 seconds until he hits rubber again.
Answer:
312,497.5Joules
Explanation:
Work done = force × distance
W = FS
Get the force
F = ma
F = 1250×9.8
F = 12250N
Get the distances using the equation of motion
v² = u² +2gS
30² =20²+2(9.8)S
900 =400+19.6S
900-400 =19.6S
500 = 19.6S
S = 500/19.6
S = 25.51m
Get the work done
Work done = 12250×25.51
Workdone = 312,497.5Joules
Describing motion from each frame of reference:
Since observer A is on one of the train's cars, he will find that he is at rest with respect to the train even when it is pulling away because, he is also moving with respect to the train.
Observer B on the stationary platform will observe that the train is pulling away from the station towards the right of the platform. as is described
Observer C will notice that the train is approaching him in the opposite direction at a speed which is the sum of the speeds at which both the train are traveling
If the conductor applies brakes on the train, since the platform is a stationary frame of reference. The motion will be observed as a simple decelerationg. he will observe that the force due to the brakes will cause the velocity of the train with respect to the train to decrease
Universal law of gravitation effects all objects alike . There will be a constant force of gravity acting on the train that will keep the train on its tracks. The tracks in turn exert a reaction force on the train. There will not be any affect on the train's motion as such( assuming that the train is moving along the tracks and gravitational force is exactly perpendicular)
