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Verizon [17]
1 year ago
5

How many air molecules are in a 13.0×12.0×10.0 ft room (28.2 L=1 ft3)? Assume atmospheric pressure of 1.00 atm, a room temperatu

re of 20.0 ∘C, and ideal behavior.
Physics
1 answer:
aksik [14]1 year ago
4 0

Answer:

1963.93 Moles

Explanation:

-We know the standard conversion ratio for the volume of a mole is 1 Mole=22.4L

Given volume of rooms as 13.0ft\times12.0ft\times10ft=1560 ft^3

Convert the volume into liters:1560\times28.2l=43992l

#From our conversion ratio above, we get the volume of air molecules in moles as:

V_m=\frac{43992L}{22.4L}\times 1\ mole\\\\=1963.93\ Moles

Hence, the volume of air molecules is 1963.93 Moles

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A spring with a spring constant of 0.70 N/m is stretched 1.5 m. What was the force?
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Answer:

1.05 N

Explanation:

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from Hooke's law of elasticity

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The statements below are all true. Some of them represent important reasons why the giant impact hypothesis for the Moon’s forma
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the order of importance must be     b e a f c

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Modern theories indicate that the moon was formed by the collision of a bad plant with the Earth during its initial cooling period, due to which part of the earth's material was volatilized and as a ring of remains that eventually consolidated in Moon.

Based on the aforementioned, let's analyze the statements in order of importance

b) True. Since the moon is material evaporated from Earth, its compassion is similar

e) True. If the moon is material volatilized from the earth it must train a finite receding speed

a) True. The solar system was full of small bodies in erratic orbits that wander between and with larger bodies

f) False. The moon's rotation and translation are equal has no relation to its formation phase

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Therefore, the order of importance must be

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5 0
2 years ago
Object A of mass M is moving east at speed v. It collides with object B of mass 2M that was initially at rest. The motion of the
Firlakuza [10]

Answer:

v_B=\frac{v}{3}

Explanation:

Given that:

mass of object A, m_A=M

mass of object B, m_B=2M

speed of object A, v_A=v

So, according to the conservation of momentum, the momentum before collision is equal to the momentum after conservation.

m_A.v+m_B\times 0=m_A\times v_A +m_B\times v_B

M\times v+0 = M\times \frac{v}{3}+2M\times v_B

2M\times v_B= \frac{2M\times v}{3}

v_B=\frac{v}{3}

3 0
2 years ago
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