Answer:
Both Thomson and Rutherford used charged particles in their experiments.
Explanation:
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
The geosphere interacts with the hydrosphere when water causes rock to erode. The atmosphere provides the geosphere with heat and energy for erosion, and the geosphere reflects the sun's energy back into the atmosphere.
Answer:
Option D. ²²²₉₀Th
Explanation:
Let the unknown be ⁿₘZ. Thus, the equation becomes:
²²⁶₉₂U —> ⁴₂He + ⁿₘZ
Next, we shall determine n, m and Z. This can be obtained as follow:
For n:
226 = 4 + n
Collect like terms
226 – 4 = n
222 = n
n = 222
For m:
92 = 2 + m
Collect like terms
92 – 2 = m
90 = m
m = 90
For Z:
ⁿₘZ => ²²²₉₀Z => ²²²₉₀Th
Therefore, the complete equation becomes:
²²⁶₉₂U —> ⁴₂He + ⁿₘZ
²²⁶₉₂U —> ⁴₂He + ²²²₉₀Th
Thus, the unknown is ²²²₉₀Th
