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3 years ago

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A projectile is launched with a velocity of 13.2m/s at an angle of 37 degrees above the horizontal. What is the horizontal compo

nent of the projectiles velocity 1s after the object is fired. PLS HURRY THIS IS TIMEd

1 answer:

MrRissso [65]3 years ago

8 0

Answer:

horizontal component = 10.54m/s

Explanation:

horizontal component = 13.2cos37°

horizontal component = 10.54m/s

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Fill in the blanks for the following:

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Answer:

<em>a. 4.21 moles</em>

<em>b. 478.6 m/s</em>

<em>c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

Explanation:

Volume of container = 100.0 L

Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

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Therefore,

n = (1.01325 x 100)/(0.08206 x 293)

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<em>For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship</em>

$\frac{Voxy}{Vnit}$ = $\sqrt{\frac{Mnit}{Moxy} }$

where

Voxy = root mean square velocity of oxygen = 478.6 m/s

Vnit = root mean square velocity of nitrogen = ?

Moxy = Molar mass of oxygen = 31.9 g/mol

Mnit = Molar mass of nitrogen = 14.00 g/mol

$\frac{478.6}{Vnit}$ = $\sqrt{\frac{14.0}{31.9} }$

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