Both valves are closed during the power stroke.
While the fuel is burning in the cylinder, you want
all the force of the expanding gases to push the
piston down ... you don't want any of the gases
or their pressure escaping.
If either of the valves was open, even just a crack,
then part of the gases would go blooey out the valve,
and some pressure would be lost that's supposed to be
pushing the piston.
The box is accelerated from rest to 4 m/s in a matter of 2.5 s, so its acceleration <em>a</em> is such that
4 m/s = <em>a</em> (2.5 s) → <em>a</em> = (4 m/s) / (2.5 s) = 1.6 m/s²
Then the force applied to the box has a magnitude <em>F</em> such that
<em>F</em> = (10 kg) (1.6 m/s²) = 16 N
Answer: 0.091 m
Explanation:
r = 1/B * √(2mV/e), where
r = radius of their circular path
B = magnitude of magnetic field = 1.29 T
m = mass of Uranium -238 ion = 238 * amu = 238 * 1.6*10^-27 kg
V = potential difference = 2.9 kV
e = charge of the Uranium -238 ion = 1.6*10^-19 C
r = 1/1.29 * √[(2 * 238 * 1.6*10^-27 * 2900) / 1.6*10^-19]
r = 1/1.29 * √(2.21*10^-21 / 1.6*10^-19)
r = 1/1.29 * √0.0138
r = 1/1.29 * 0.117
r = 0.091 m
Therefore, the radius of their circular path is 0.091 m
