The one that does that is the photoelectric effect. This allows the sun to hit the surface and produces enough energy for electrons to be knocked off the atom and allows a current to move.
The product of speed and time is distance. To calculate the total distance you multiple the speed in kilometers per second by the time at that speed in seconds, do this for all 3 different speeds then add them up, the 17.4 minutes eating does not affect the answer at all. to convert from minutes to seconds multiply time in minutes by 60, to convert from km/h to km/s divide km/h by 3600.
(23.5x60)x(74.5/3600) = 29.2km (rounded to 1 decimal place)
+
(15.9x60)x(111/3600) = 29.4km (rounded to 1 decimal place)
+
(49.2x60)x(38.7/3600) = 31.7km
=90.3km
The brackets are not necessary but i think it makes it more clear what is happening in your working.
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:
4.13Hz
Explanation:
f1 = 1/t1 = 1/0.022 = 45.45 Hz
f2 = 1/t2 = 1/0.0242= 41.32 Hz
No. of beats
= 45.45- 41.32
~ 4.13Hz
Answer:
M g H = 1/2 M v^2 potential energy = kinetic energy
v^2 = 2 g H = 2 * 9.80 * 6 = 117.6 m/s^2
v = 10.8 m/s
(C)
