Answer:
Air resistance have some significant role in projectile motion if the motion lasts for some time.
Explanation:
- Air resistance or air drag seems to be important in daily actvities and games like baseball.
- The trajectory of the projectile with or without air resistance or air drag is totally different.
- When we neglect air drag, the only acting force is gravity against the motion so the maximum height and range are suppose Hmax and R.
- Now, when we consider air drag, it is important to notice that there are two forces against the motion of the ball and along the direction of gravity. It seems that both maximum height and range are lesser Hmax'< Hmax and R'<R.
I would rather be hit by the deflated ball because it wouldn't hurt as bad because it wouldn't have a lot of weight to hurt me in anyway
Answer:
The value of tangential acceleration 
40 
The value of radial acceleration 
Explanation:
Angular acceleration = 50 
Radius of the disk = 0.8 m
Angular velocity = 10 
We know that tangential acceleration is given by the formula 
Where r = radius of the disk
= angular acceleration
⇒ 0.8 × 50
⇒ 40
This is the value of tangential acceleration.
Radial acceleration is given by
Where V = velocity of the disk = r 
⇒ V = 0.8 × 10
⇒ V = 8 
Radial acceleration
This is the value of radial acceleration.
“Charged objects have an imbalance of charge - either more negative electrons than positive protons or vice versa. And neutral objects have a balance of charge - equal numbers of protons and electrons. The principle stated earlier for atoms can be applied to objects. Objects with more electrons than protons are charged negatively; objects with fewer electrons than protons are charged positively.
In this discussion of electrically charged versus electrically neutral objects, the neutron has been neglected. Neutrons, being electrically neutral play no role in this unit. Their presence (or absence) will have no direct bearing upon whether an object is charged or uncharged. Their role in the atom is merely to provide stability to the nucleus.”
Hope this helps a bit.
!! (Credits to The Psychics Classroom) !!
Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.
Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:
Pcos 15°-N=0
Psin15°-f= m*ac
from the first we obtain N, the normal force
N=750Kg*9.8* cos (15°)= 7.1 *10^3 N
Then to calculate the frictional force (f) we can use the second equation
f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r
f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N
