Answer:
a) 0 J
b) W = nRTln(Vf/Vi)
c) ΔQ = nRTln(Vf/Vi)
d) ΔQ = W
Explanation:
a) To find the change in the internal energy you use the 1st law of thermodynamics:
Q: heat transfer
W: work done by the gas
The gas is compressed isothermally, then, there is no change in the internal energy and you have
ΔU = 0 J
b) The work is done by the gas, not over the gas.
The work is given by the following formula:
n: moles
R: ideal gas constant
T: constant temperature
Vf: final volume
Vi: initial volume
Vf < Vi, then W < 0 and the work is done on the gas
c) The gas has been compressed. Thus, its temperature increases and heat has been transferred to the gas.
The amount of heat is equal to the work done W
d)
Answer:
6.378 * 10^6 m
Explanation:
Radius = 6378 km
Changing in metres
6378 km = 6378000 m
Now in scientific notation
= 6.378 * 10 ^ 6 m
Answer:
I wish i saw the field shown to the right
Explanation:
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.
Answer:
a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,
v/c% = 2.33 10⁻²
Explanation:
a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy
starting point. Where the electrons come out
Em₀ = U = e DV
final point. Where they hit the target
Em_f = K = ½ m v2
energy is conserved
Em₀ = Em_f
e ΔV = ½ m v²
ΔV = 
mv²/e (1)
If the speed of light is c and this is 100% then 1% is
v = 1% c = c / 100
v = 3 10⁸/100 = 3 10⁶6 m/ s
let's calculate
ΔV = 
ΔV = 25.59 V
b) Ask for the potential difference for protons with the same kinetic energy as electrons
K_p = ½ m v_e²
K_p = 9.1 10⁻³¹ (3 10⁶)²
K_p = 40.95 10⁻¹⁹ J
we substitute in equation 1
ΔV = Kp / M
ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹
ΔV = 25.59 V
notice that these protons go much slower than electrons because their mass is greater
c) The speed of the protons is
e ΔV = ½ M v²
v² = 2 e ΔV / M
v² = 
v² = 49,035 10⁸
v = 7 10⁴ m / s
Relation
v/c = 
v/c= 2.33 10⁻⁴
