Answer:
These regions are known as compressions and rarefactions respectively. The compressions are regions of high air pressure while the rarefactions are regions of low air pressure.
Explanation:
Answer:
F = 51.3°
Explanation:
The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:
![F = Parallel\ Component\ of\ Weight\ of\ Wagon= WSin\theta\\](https://tex.z-dn.net/?f=F%20%3D%20Parallel%5C%20Component%5C%20of%5C%20Weight%5C%20of%5C%20Wagon%3D%20WSin%5Ctheta%5C%5C)
where,
W = Weight of Wagon = 150 N
θ = Angle of Inclinition = 20°
Therefore,
![F = (150\ N)Sin\ 20^o](https://tex.z-dn.net/?f=F%20%3D%20%28150%5C%20N%29Sin%5C%2020%5Eo)
<u>F = 51.3°</u>
Mass have no effect for the projectile motion and u want to know the height "h"
first,
find the vertical and horizontal components of velocity
vertical component of velocity = 12 sin 61
horizontal component of velocity = 12 cos 61
now for the vertical motion ;
S = ut + (1/2) at^2
where
s = h
u = initial vertical component of velocity
t = 0.473 s
a = gravitational deceleration (-g) = -9.8 m/s^2
h=[12×sin 610×0.473]+[−9.8×(0.473)2]
u can simplify this and u will get the answer
h=.5Gt2
H=1.09m
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill:
![v=15 m/s](https://tex.z-dn.net/?f=v%3D15%20m%2Fs)
- radius of the hill:
![r=100 m](https://tex.z-dn.net/?f=r%3D100%20m)
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car
![W=mg](https://tex.z-dn.net/?f=W%3Dmg)
(downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force,
![m \frac{v^2}{r}](https://tex.z-dn.net/?f=m%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20)
, so we can write:
![mg-N=m \frac{v^2}{r}](https://tex.z-dn.net/?f=mg-N%3Dm%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20)
(1)
By rearranging the equation and substituting the numbers, we find N:
![N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N](https://tex.z-dn.net/?f=N%3Dmg-m%20%5Cfrac%7Bv%5E2%7D%7Br%7D%3D%28975%20kg%29%289.81%20m%2Fs%5E2%29-%28975%20kg%29%20%5Cfrac%7B%2815%20m%2Fs%29%5E2%7D%7B100%20m%7D%3D7371%20N%20%20)
(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
![N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N](https://tex.z-dn.net/?f=N%3Dmg-m%20%5Cfrac%7Bv%5E2%7D%7Br%7D%3D%2862%20kg%29%289.81%20m%2Fs%5E2%29-%2862%20kg%29%20%5Cfrac%7B%2815%20m%2Fs%29%5E2%7D%7B100%20m%7D%3D469%20N%20)
(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
![mg=m \frac{v^2}{r}](https://tex.z-dn.net/?f=mg%3Dm%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20)
from which we find