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
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Answer:
Distance = 3.69 × 10^9 m
The distance from the probe to Earth is 3.69 × 10^9 m
Explanation:
Distance from the probe to the Earth can be derived using the simple motion formula;
Distance = speed × time .....1
Since a radio signal uses an electromagnetic wave to transfer signal, it has the same speed as the speed of light.
Speed of radio signal = speed of light = 3.0 × 10^8 m/s
time taken to reach the earth = 12.3 seconds
Substituting the values of speed and time into equation 1;
Distance = 3.0 × 10^8 m/s × 12.3 s
Distance = 36.9 × 10^8 m
Distance = 3.69 × 10^9 m
Note: all electromagnetic radiation have the same speed which is equal to 3.0 × 10^8 m/s
The mass of an atom comes from the protons and neutrons that is found in the nucleus. The number of protons is the atomic number of an element. To find the number of neutrons, subtract the atomic number from the mass of an atom. For example, sodium’s atomic number is 11. This will tell us that sodium has 11 protons in it. The atomic mass of sodium is 23. So subtract 23 form 11 gives us 12. Therefore, there are 12 neutrons in sodium.
