Answer:
The particle which completes the given equation is :
Explanation:
The given reaction is of a fission reaction:
Total mass on the reactant side is equal to the total mass on the product side:
239 + 1 = 100 +A+ 2
A = 138
Sum of atomic numbers on the reactant side is equal to the sum of atomic number on the product side:
94 + 1(0) = 40 + Z + 2(0)
Z = 54
So atomic number 54 id of Xenon.
The particle which completes the given equation is :
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
Just find the density of every metal and select the one with a density of 2.71 g/cm³ . This is:
Metal 1
ρ = m/V
ρ = 22.1 g / 3 cm³
ρ = 7.367 g / cm³
Metal 2
ρ = m/V
ρ = 42 g / 4 cm³
ρ = 10.5 g / cm³
Metal 3
ρ = m/V
ρ = 9.32 g / 5 cm³
ρ = 1.864 g / cm³
Metal 4
ρ = m/V
ρ = 8.13 g / 3 cm³
ρ = 2.71 g / cm³
<h2>R / Metal 4 was selected.</h2>
