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Jet001 [13]
2 years ago
8

What is the average velocity of atoms in 2.00 mol of neon (a monatomic gas)

Physics
2 answers:
Dimas [21]2 years ago
8 0

Answer:

v = 876 m/s

Explanation:

It is given that,

Number of mol of Neon is 2 mol

Temperature, T = 308 K

Mass, m = 0.02 kg

Value of R - 8.31 J/mol-K

We need to find the average velocity of atoms in 2.00 mol of neon. Neon is a monoatomic gas. Let v is the velocity. So,

\dfrac{1}{2}mv^2=\dfrac{3}{2}nRT\\\\v=\sqrt{\dfrac{3nRT}{m}} \\\\v=\sqrt{\dfrac{3\times 2\times 8.314\times 308}{0.02}} \\\\v=876.47\ m/s

So, the correct option is (B).

Alenkasestr [34]2 years ago
5 0
I think it is 887m/s I hope this helps if not I’m really sry
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Some expressways are curved, banked and designed to maximize ___________ at higher speeds.
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Answer:

Safety

Explanation:

Expressways are banked to resist centifugal action

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Vector vector a has a magnitude of 29 units and points in the positive y-direction. when vector vector b is added to vector a ,
Nutka1998 [239]
Good morning.

We have:

\mathsf{\overset{\to}{a} = 29\overset{\to}{j}}

Where j is the unitary vector in the direction of the y-axis.

We have that 

\mathsf{\overset{\to}{a}+\overset{\to}{b} = -18\overset{\to}{j}}

We add the vector -a to both sides:

\mathsf{\overset{\to}{b} = -18\overset{\to}{j} -\overset{\to}{a} = -18\overset{\to}{j} -29\overset{\to}{j}}\\ \\ \mathsf{\overset{\to}{b}=-47\overset{\to}{j}}


Therefore, the magnitude of b is 47 units.
5 0
2 years ago
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the jet plane travels along the vertical parabolic path. when it is at point a it has speed of 200 m/s, which is increasing at t
givi [52]

Explanation:

Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.

→ The tangential component of acceleration is rate of increase in the speed of plane so,

a_{t} = v = 0.8 m/s^{2}

→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).

dy/dx = d(0.4x²)/dx

         = 0.8x

Take the derivative again,

d²y/dx² = d(0.8x)/dx

          = 0.8

at x= 5 Km

dy/dx = 0.8(5)

         = 4

p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} }   }{\frac{d^{2y} }{dx^{2} } }

now insert the values,

p = \frac{[1+(4)^{2}]^{\frac{3}{2} }  }{0.8}  = 87.62 km

→ Now the normal component of acceleration is given by

a_{n} = \frac{v^{2} }{p}

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aₙ = 0.457 m/s²

→ Now the total acceleration is,

a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}

a = [(0.8)^{2} + (0.457)^{2}]^{0.5}

a = 0.921 m/s²

4 0
2 years ago
Lori’s family is on a road trip. They split their drive into the five legs listed in the table. Find the average velocity for ea
NARA [144]
It's not possible to answer the question exactly the way it's written.
That's because we don't know anything about the direction they
drive at any time during the trip. 

You see, "velocity" is not just a word that you use for 'speed' when
you want to sound smart and technical, like this question is doing. 
"Velocity" is a quantity that's made up of speed AND THE DIRECTION
of the motion.  If you don't know the direction of the motion, then you
CAN'T tell the velocity, only the speed.

Here are the average speeds that Lori's family drove on each leg
of their trip:

Speed = (distance covered) / (time to cover the distance) .

Leg-A:  
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Leg-B:  
Speed = 20km/15min = (1 and 1/3) km/min

Leg-C  
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Leg-D:  
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Leg-E: 
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From lowest speed to highest speed, they line up like this:

[Leg-E] ==> [Leg-B] ==> [Leg-A] ==> [Leg-C] ==> [Leg-D]
  1.0 . . . . . . . . 1.3 . . . . . . . 1.5 . . . . . . . 2.0 . . . . . . . 4.0 . . . . km/minute   

Whoever drove Leg-D should have been roundly chastised
and then abandoned by the rest of the family.  36 km in 9 minutes
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Answer:

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