The solution is:A mole of gas occupies 22.4 L A liter is 1000 cubic centimeters because 1 cubic centimeter = 1 mL
One Helium molecule (essentially one helium atom) has atomic mass 4 g/mol So for every 22400 cubic centimeters of volume, we have 4 grams of helium Density of helium = 4g / 22400 cm^3 = 1g / 5600 cm^3
Volume of a sphere = (4/3)(pi)r^3 Volume of the outside sphere (the entire sphere) is (4/3)(pi)(R+T)^3 Volume of the inside sphere (the hollow region) is (4/3)(pi)R^3
The difference for the volume of silver. (4/3)(pi)(R+T)^3 - (4/3)(pi)R^3 = (4/3)(pi)(3R^2T + 3RT^2 + T^3)
The density of silver is 10.5g/cm^3
So the mass of the silver is computed by:10.5*(4/3)(pi)(3R^2T + 3RT^2 + T^3) = (14*pi)(3R^2T + 3RT^2 + T^3) = (14pi)T(3R^2 + 3RT + T^2)
Now for the mass of helium: volume x density = (4/3)(pi)R^3 (1/5600) = (pi/4200)R^3
Set the two masses equal: (pi/4200)R^3 = (14pi)T(3R^2 + 3RT + T^2) R^3 = 58800*T(3R^2 + 3RT + T^2) R / T = 58800*(3R^2 + 3RT + T^2) / R^2 = 58800*( 3+T/R^2+(T/R)^2)
then solve for xx = T / R 1/x = 58800*( 3+x/R+x^2) 1/58800 = x (3 + x/R + x^2) 1/58800 = 3x + x^3 x^3 + 3x - 1/58800= 0 x = ~ 5.66893x10^(-6)
During week 2, he could type 15 words per minute, at his fourth week, he could type 30. So 30-15=15. He could type 15 words faster on week 4 than he could at week 2.
Answer:
1 would: You dont have to pay anything because 80% of 750 is $150 which you also have to deduct $150 so you dont pay anything.
2: 90% of $575 is $517 so 58 dollars would be deducted.
Hope this helps:)
Let r be a radius of a given circle and α be an angle, that corresponds to a sector.
The circle area is
and denote the sector area as
.
Then
(the ratio between area is the same as the ratio between coresponding angles).
.
Answer:
(±√3,±√+1)
Step-by-step explanation:
