Kepler’s
third law formula: T^2=4pi^2*r^3/(GM)
We’re trying to find M, so:
M=4pi^2*r^3/(G*T^2)
M=4pi^2*(1.496
× 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)
M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))
Let’s work
with the units:
(m^3)/((N*m^2/kg^2)*days^2))=
=(m^3*kg^2)/(N*m^2*days^2)
=(m*kg^2)/(N*days^2)
=(m*kg^2)/((kg*m/s^2)*days^2)
=(kg)/(days^2/s^2)
=(kg*s^2)/(days^2)
So:
M=1.48× 10^40(kg*s^2)/(days^2)
Now we need to convert days to seconds in order to cancel
them:
1 day=24 hours=24*60minutes=24*60*60s=86400s
M=1.48× 10^40(kg*s^2)/((86400s)^2)
M=1.48× 10^40(kg*s^2)/(
86400^2*s^2)
M=1.48× 10^40kg/86400^2
M=1.98x10^30kg
The
closest answer is 1.99
× 10^30
(it may vary
a little with rounding – the difference is less than 1%)
<span>3.36x10^5 Pascals
The ideal gas law is
PV=nRT
where
P = Pressure
V = Volume
n = number of moles of gas particles
R = Ideal gas constant
T = Absolute temperature
Since n and R will remain constant, let's divide both sides of the equation by T, getting
PV=nRT
PV/T=nR
Since the initial value of PV/T will be equal to the final value of PV/T let's set them equal to each other with the equation
P1V1/T1 = P2V2/T2
where
P1, V1, T1 = Initial pressure, volume, temperature
P2, V2, T2 = Final pressure, volume, temperature
Now convert the temperatures to absolute temperature by adding 273.15 to both of them.
T1 = 27 + 273.15 = 300.15
T2 = 157 + 273.15 = 430.15
Substitute the known values into the equation
1.5E5*0.75/300.15 = P2*0.48/430.15
And solve for P2
1.5E5*0.75/300.15 = P2*0.48/430.15
430.15 * 1.5E5*0.75/300.15 = P2*0.48
64522500*0.75/300.15 = P2*0.48
48391875/300.15 = P2*0.48
161225.6372 = P2*0.48
161225.6372/0.48 = P2
335886.7441 = P2
Rounding to 3 significant figures gives 3.36x10^5 Pascals.
(technically, I should round to 2 significant figures for the result of 3.4x10^5 Pascals, but given the precision of the volumes, I suspect that the extra 0 in the initial pressure was accidentally omitted. It should have been 1.50e5 instead of 1.5e5).</span>
Answer:
θ = 66.90°
Explanation:
we know that
I= intensity of polarized light =1
I_o= intensity of unpolarized light = 13
putting vales we get
⇒
therefore θ = 66.90°
