Answer:
114 kPa
Explanation:
By Bernoulli's equation when a fluid flows steadily through a pipe:
P + ρ*g*y + v² = constant in the pipe, where P is the pressure, ρ is the density of the fluid, g is the gravity acceleration (9.8 m/s²), y is the high, and v the velocity.
By the continuity equation, the liquid flow must be constant in the pipe, and then:
A1*v1 = A2*v2
Where A is the area, v is the velocity, 1 is the point 1, and 2 the point 2 in the pipe. The are is the circle area: π*(d/2)². So:
π*(0.105/2)²*9.91 = π*(0.167/2)²*v2
0.007v2 = 0.027
v2 = 3.9 m/s
Then:
P1 + ρ*g*y1 + v1² = P2 + ρ*g*y2 + v2²
ρ*g*y1 - ρ*g*y2 + v1² - v2² = P2 - P1
ρ*g*Δy + v1² - v2² = ΔP
ΔP = 1290*9.8*9.01 + 9.91² - 3.9²
ΔP = 113,987.42 Pa
ΔP = 114 kPa
If Thomson’s model of the atom were correct, Rutherford should have seen a complete reflection of those particles when fired at the foil. Those particles would have hit a solid object (Thomson model) and would have been reflected back at the emitter. However, Rutherford discovered that most of those particles passed through the sheet, either continuing straight through or having the angle change a small amount (bouncing off the nucleus). Only a small number of the particles were reflected back, having hit the nucleus straight on. This led him to believe that there was a small, dense portion of the atom that resided within a larger, emptier space.
Answer:
There is at least 2 ways. First, the isotopes are allowed to heat water into steam, and the steam turns a generator. 2nd, the isotopes are allowed to heat one end of a "thermopile", and the thermopile powers electronics
Hope this helped you!
