A molecule of ethanol has two carbon atoms, six hydrogen atoms, and one oxygen atom. A ball-and-stick model of a molecule of eth
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<u>Answer:</u>
Maximum height reached = 35.15 meter.
<u>Explanation:</u>
Projectile motion has two types of motion Horizontal and Vertical motion.
Vertical motion:
We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
Considering upward vertical motion of projectile.
In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g and final velocity = 0 m/s.
0 = u sin θ - gt
t = u sin θ/g
Total time for vertical motion is two times time taken for upward vertical motion of projectile.
So total travel time of projectile = 2u sin θ/g
Horizontal motion:
We have equation of motion , , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 and time taken = 2u sin θ /g
So range of projectile,
Vertical motion (Maximum height reached, H) :
We have equation of motion, , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.
Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H
In the give problem we have R = 301.5 m, θ = 25° we need to find H.
So
Now we have
So maximum height reached = 35.15 meter.
1.
For the first part, we just need to write the equation of the forces along two perpendicular directions.
We have actually only two forces acting on the car, if we want it to go around the track without friction:
- The weight of the car, mg, downward
- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)
By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:
(1)
(2)
where in the second equation, the term represents the centripetal force, with v being the speed of the car and R the radius of the track.
Dividing eq.(2) by eq.(1), we get the following expression:
2.
In this second situation, the cars moves around the track at a speed
This means that the centripetal force term
is now larger than before, and therefore, the horizontal component of the normal reaction, , is no longer enough to keep the car in circular motion.
This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes
And re-arranging for F,
(3)
But from eq.(2) in the previous part we know that
So, susbtituting into eq.(3),