Answer:
The ocean tides on earth are caused by both the moon's gravity and the sun's gravity. ... Even though the sun is much more massive and therefore has stronger overall gravity than the moon, the moon is closer to the earth so that its gravitational gradient is stronger than that of the sun.
Answer:
the work is done by the gas on the environment -is W= - 3534.94 J (since the initial pressure is lower than the atmospheric pressure , it needs external work to expand)
Explanation:
assuming ideal gas behaviour of the gas , the equation for ideal gas is
P*V=n*R*T
where
P = absolute pressure
V= volume
T= absolute temperature
n= number of moles of gas
R= ideal gas constant = 8.314 J/mol K
P=n*R*T/V
the work that is done by the gas is calculated through
W=∫pdV= ∫ (n*R*T/V) dV
for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:
W=∫pdV= ∫ (n*R*T/V) dV = n*R*T ∫(1/V) dV = n*R*T * ln (V₂/V₁)
since
P₁=n*R*T/V₁
P₂=n*R*T/V₂
dividing both equations
V₂/V₁ = P₁/P₂
W= n*R*T * ln (V₂/V₁) = n*R*T * ln (P₁/P₂ )
replacing values
P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa
since P₂ = 1 atm = 101325 Pa
W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J
Answer: 20.4752789138x x 10^23 atoms
To count how many atoms in moles you need to know Avogadro's number. Avogadro's number dictate that for every mole there is 6.022140857 × 10^23 molecule/atoms.
Then 3.4 moles of helium will be 3.4x 6.022140857 x 10^23 atoms= 20.4752789138x x 10^23 atoms
The question is incomplete, the options are;
RI^2
I^2/R
R/I^2
R/V^2
RV^2
V^2/R
VI
VIR
Select all that apply
Answer:
P=RI^2
P=V^2/R
P=VI
Explanation:
Power is the rate at which energy is changing in a circuit. It is shown by the formulas outlined above from the group of answer choices. Since the current (I), voltage (V), and resistance (R) were mentioned in the question, any of three three formulas could be used to obtain the power drawn by the conductor.
