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Montano1993 [528]
3 years ago
9

You have four fixed-volume containers at STP . Container A has 0.5 mol of gas in 11.2 L. Container B has 2 mol of gas in 22.4 L.

Container C has 1 mol of gas in 22.4 L. Container D has 2 mol of gas in 11.2 L. Which containers have equal pressures?
Physics
1 answer:
Colt1911 [192]3 years ago
8 0

Answer:

Container A and C

Explanation:

ideal gas equation gives P=nRT/V

so at constant Temperature and pressure, P=n/T

Container A and C after dividing number of moles and Volume, are found to be the same=0.0446

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A cyclist is riding a bicycle at a speed of 22 mph on a horizontal road. The distance between the axles is 42 in., and the mass
stealth61 [152]

Answer:

The shortest distance is  S = 24.86 ft

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

   The speed of the bicycle is v_b = 22\ mph = 22 * \frac{5280}{3600}   =  32.26 ft/s

     The distance between the axial is  d = 42 \ in

The mass center of the cyclist and the bicycle is m_c = 26 \ in  behind the front axle

The mass center of the cyclist and the bicycle is m_h = 40 \ in above the ground

   For the bicycle not to be thrown over the

     Momentum about the back wheel must be zero so

                \sum _B = 0

=>             mg (26) = ma(40)

=>             a = \frac{26}{40} g

Here  g = 32.2 ft/s^2

     So     a =  \frac{26}{40} (32.2)

             a =  20.93 ft/s^2

Apply the equation of motion to this motion we have

       v^2 = u^2 + 2as

 Where  u = 32.26 ft /s

             and v = 0 since the bicycle is coming to a stop

        v^2 = (32.26)^2 - 2(20.93) S

=>      S = 24.86 ft        

                 

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3 years ago
The first impression you give is the most important what does it mean?
Neko [114]

Answer:

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What does FCCLA stand for?
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alexandr1967 [171]
Rate of speed (3 m/s north is three miles per second north, so it's a rate of speed)
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3 years ago
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A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 4.00 m at constant speed. If the coeff
zysi [14]

Answer:

Work done is equal to 125.44 J

Explanation:

We have given mass of the student m = 80 kg

Distance moved d = 4 m

Acceleration due to gravity g=9.8m/sec^2

Coefficient of kinetic friction \mu =0.4

So normal force exerted by student F=\mu mg=0.4\times 80\times 9.8=313.6N

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