The instrument is a Geiger counter and is used to measure radioactive level around people's bodies.
Answer:
The value is 
Explanation:
From the question we are told that
The speed of the rope with hook is 
The angle is 
The speed at which it hits top of the wall is 
Generally from kinematic equation we have that
Here h is the height of the wall so
![[16.3 sin (65)]^2 = [24.1 sin (65)] ^2+ 2 (-9.8)* h](https://tex.z-dn.net/?f=%5B16.3%20sin%20%2865%29%5D%5E2%20%3D%20%20%5B24.1%20sin%20%2865%29%5D%20%5E2%2B%20%20%202%20%28-9.8%29%2A%20h)
=>
Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
Answer:
A hypothesis is what you think will happen.
A conclusion is the results of an experiment summarized.
Hope this helps.
