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faust18 [17]
3 years ago
6

Calculate the density of helium gas at 25.0â—¦c and at a pressure of 1521 mmhg. express your answer in g/l.

Physics
1 answer:
LuckyWell [14K]3 years ago
4 0
Ideal gas equation: pV = nRT

Transform it to show density:

pV = [mass/Molar mass] RT

mass/V = (p*molar mass) /(RT)

density = mass / V

density = p*molar mass / (RT)

Now susbtitute the data

p = 1521 mmHg * 1 atm / 760 mmHg = 2.0 atm
T = 25 °C + 273.15 = 298.15 k
molar mass of He = 4.0 g/mol
R = universal gas constant = 0.082 atm*L / mol*k

density =2.0 atm*4.0 g/mol / (0.082 atm*L / mol*k *298.15k)

density = 0.327 g/L
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Answer:

100 m/s

Explanation:

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Fx = 100 * cos (37)   = 79.86 m/s

Fy = 100 * sin (37)   =  60.18 m/s

applying motion equations

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assuming no acceleration in horizontal direction we can assure that the ending velocity component will still remain  Fx = 79.86 m/s

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V² = U² +2as

    =  60.18² + 2*(-10)*0

Fy  = 60.18 m/s

Total speed = √(F²x+F²y) = 100 m /s

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3 years ago
An SUV has a mass of 6000 kg and a compact car has a mass of 2000 kg. They both go around the same unbanked circular curve on th
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Answer:

Therefore, the centripetal acceleration of the SUV is <u>3 times</u> ao of the compact car.

Explanation:

FOR SUV:

ac_{suv} = \frac{mv^2}{r}\\

where,

ac_suv = centripetal acceleration of the SUV = ?

m = mass of SUV = 6000 kg

v = speed of SUV = vo

r = radius of path

Therefore,

ac_{suv} = \frac{6000\ vo^2}{r}\\---------------- equation (1)

FOR CAR:

ao = \frac{mv^2}{r}\\

where,

ao = centripetal acceleration of the car = ?

m = mass of car = 2000 kg

v = speed of car = vo

r = radius of path

Therefore,

ao = \frac{2000\ vo^2}{r}\\--------------------- equation (2)

Dividing equation (1) by eq(2):

\frac{ac_{suv}}{ao} = \frac{6000}{2000}\\\\ac_{surv} = 3ao

Therefore, the centripetal acceleration of the SUV is <u>3 times</u> ao of the compact car.

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