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:
3.0 x10^-3 J
Explanation:
The potential energy of a spring is given by PE = (0.5)k*x^2
Where
K: Spring Constant = 60 N/m
x: displacement of the spring from its equilibrium position = 1cm = 0.01m
Then PE = 0.5(60)(.01)^2 = 0.003J = 3.0 x10^-3 J
Answer:
Total charge provided by the battery could be 900000 C.
Maximum current provided by the battery for 37 minutes could be 405.405 A
Explanation:
Rating= 250 A-h
a. Total charge:
Suppose t=1h
We konw that 
, replacing:
Total charge provided by the battery could be 900000 C.
b. Maximum current for 37 minutes
Maximum current provided by the battery for 37 minutes could be 405.405 A
See from periodic table the proton or atomic number of elements u want to know about then as u know first electron shell can hold 2 electrons 2nd can hold eight as well as all others........protons are equal to electrons so divide proton number into shells but remember to use amounts which it can hold
CHEERS !!
