Answer:
the buoyant force on the chamber is F = 7000460 N
Explanation:
the buoyant force on the chamber is equal to the weight of the displaced volume of sea water due to the presence of the chamber.
Since the chamber is completely covered by water, it displaces a volume equal to its spherical volume
mass of water displaced = density of seawater * volume displaced
m= d * V , V = 4/3π* Rext³
the buoyant force is the weight of this volume of seawater
F = m * g = d * 4/3π* Rext³ * g
replacing values
F = 1025 kg/m³ * 4/3π * (5.5m)³ * 9.8m/s² = 7000460 N
Note:
when occupied the tension force on the cable is
T = F buoyant - F weight of chamber = 7000460 N - 87600 kg*9.8 m/s² = 6141980 N
Static Friction
It is the friction that exists between a stationary object and the surface on which it's resting.
Sliding friction
It is the resistance created by two objects sliding against each other.
Rolling friction:-
It is the force resisting the motion when a body rolls on a surface.
hope this helps x
Answer:
The required angle is (90-25)° = 65°
Explanation:
The given motion is an example of projectile motion.
Let 'v' be the initial velocity and '∅' be the angle of projection.
Let 't' be the time taken for complete motion.
Let 'g' be the acceleration due to gravity
Taking components of velocity in horizontal(x) and vertical(y) direction.
= v cos(∅)
= v sin(∅)
We know that for a projectile motion,
t =
Since there is no force acting on the golf ball in horizonal direction.
Total distance(d) covered in horizontal direction is -
d = ×t = vcos(∅)× = 
.
If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -
α +β = 90° as-
d = 
= ![\frac{v^{2}sin(2[90-β]) }{g}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%5E%7B2%7Dsin%282%5B90-%CE%B2%5D%29%20%7D%7Bg%7D)
=
= 
.
∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.
∴ If α = 25° , then
β = 90-25 = 65°
∴ The required angle is 65°.
This is a "trick" question.
If the elevator is traveling at constant speed, it means it is at rest. This means anything inside the elevator traveling at constant speed, weights the same as in an elevator not moving -also at rest-.
So the 100N weight's weight doesn't change in an elevator traveling at constant speed.
Use the impulse-momentum theorem.
Substitute your known values:
Hope this helps!
