Answer: The height of the fluid rise is 0.01m
Explanation:
Using the equation
h = (2TcosѲ )/rpg
h= height of the fluid rise
diameter of the tube =3mm
radius of the tube= 3/2 =1.5mm=0.0015
T= surface tension = 600mN/m=0.6N/m
Ѳ = contact angle = 
C
p= density =3.7g/cm3= 3700kg/m3
g= acceleration due to gravity =9.8m/s2
h = ( 2*0.6*0.5)/(0.0015*3700*9.8)
h = 0.6/54.39
h= 0.01m
Therefore,the height of the fluid rise is 0.01m
Rocks and sediments I believe
Answer:
The helicopter was deformed and destroyed in the inelastic collision.
Explanation:
- When two object collide there exist two way of colliding: elastic collision and inelastic collision.
- Two terms are considered during the collision: kinetic energy and momentum.
- If both of these terms are conserved in any collision then there is no significant loss of property, this is called as elastic collision.
- If only momentum is conserved but kinetic energy is converted into other forms then it is inelastic collision. In inelastic collision, the energy is lost in the form of vibration, sound etc. causing the damage to colliding object.
- Hence the deformation of helicopter was due to inelastic collision.
Answer:
32 m/s
Explanation:
The speed of a bus is 30 m/s due East wrt the passenger
He also sees a passenger on the bus walking to the back at 2 m/s.
We need to find the passenger's velocity relative to the bus. As the observer sees that the bus and the passenger are moving in opposite direction. Let v is the relative velocity. So,
v = 30 m/s + 2 m/s
v = 32 m/s
Hence, the passenger's velocity relative to the bus is 32 m/s.
A) The acceleration is due to gravity at any given point if you look at it vertically, so
.
b)
, so
. We use
and then the final speed must be 0 because it stops at the highest point. So
. Solve for
and you get
c)
, and then we plug the values:
and we already have the time from "b)", so
![Y_m_a_x = [(32sin(25))*(32sin(25)/10)] - 5(32sin(25)/10)^2](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%20%5B%2832sin%2825%29%29%2A%2832sin%2825%29%2F10%29%5D%20-%205%2832sin%2825%29%2F10%29%5E2)
; then we just rearrange it
![Y_m_a_x = 10[(32sin(25))^2/100] - 5 [(32sin(25))^2/100]](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%2010%5B%2832sin%2825%29%29%5E2%2F100%5D%20-%205%20%5B%2832sin%2825%29%29%5E2%2F100%5D%20)
and finally
![Y_m_a_x = 5[(32sin(25))^2/100] = (32sin(25))^2/20](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%205%5B%2832sin%2825%29%29%5E2%2F100%5D%20%3D%20%2832sin%2825%29%29%5E2%2F20)
