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Andrei [34K]
3 years ago
7

A gas has an initial volume of 21.7 L at a pressure of 0.4 atm and a temperature of 252 K. The pressure of the gas increases to

2 atm as the temperature increases to 323 K. What is the final volume of the gas?
Physics
2 answers:
elena-s [515]3 years ago
7 0
Theeeeeeeeeeeeeeeee answer is 5.6 L
Sloan [31]3 years ago
3 0
According to the combined gas law: (P₁*V₁)/T₁=(P₂*V₂)/T₂, where P is the pressure, V is the volume and T is the temperature.

P₁=0.4 atm = 40530 Pa
V₁=21.7 L = 0.0217 m³
T₁=252 K
P₂=2 atm = 202650 Pa
T₂=323 K
V₂=?

We need to solve for V₂:

(P₁*V₁)/T₁=(P₂*V₂)/T₂

(T₂*P₁*V₁)/T₁=P₂*V₂

(T₂*P₁*V₁)/(T₁*P₂)=V₂, now we plug in the numbers and get:

V₂=0.00556 m³, and we can see that with increased temperature and pressure the volume V₂ has decreased. 

When we transfer V₂=0.00556 m³ to liters we get that 0.00556 m³=5.56 L. That can be rounded up to V₂=5.6 L.
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A gardener pushes a 20 kg lawnmower whose handle is tilted up 37∘ above horizontal. The lawnmower's coefficient of rolling frict
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Answer:

58.4 W

Explanation:

The speed of the lawnmower is constant: this means that its acceleration is zero, so the net force on it is zero.

The equation of the forces along the two directions therefore are:

- Perpendicular to the floor: F sin \theta + R -mg =0

- Parallel to the floor: F cos \theta - \mu R = 0

where

F is the push of the gardener

R is the normal reaction

m = 20 kg is the mass

g = 9.8 m/s^2 is the acceleration of gravity

\mu=0.18 is the coefficient of friction

\theta=37^{\circ}

Solving for R,

R=mg-Fsin \theta

Substituting into the other equation,

F cos \theta - \mu (mg-Fsin \theta) = 0\\F cos \theta - \mu mg + \mu F sin \theta = 0\\F(cos \theta+\mu sin \theta)=\mu mg\\F=\frac{\mu mg}{cos \theta + \mu sin \theta}=\frac{(0.18)(20)(9.8)}{cos 37^{\circ}+0.18(sin 37^{\circ})}=38.9 N

And the power he must supply therefore is the product of this force and the speed:

P=Fv=(38.9)(1.5)=58.4 W

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Does an electrochemical cell use two terminals or a voltmeter?
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A voltmeter does so.

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A block is pushed across a rough horizontal surface from point A to point B by a force P. The magnitude of the force of friction
DochEvi [55]

Answer:

The workdone by P is W_p=2.2J

Explanation:

Generally workdone is mathematically represented as

              W = F *d

Where F is the force and d is the distance

  The total workdone = (Workdone by Force P + Workdone by frictional force )

The difference in kinetic energy

                \Delta KE  = KE_2 - KE_1

Substituting 5.6 J for KE_1 and 4.0 J for KE_2

              \Delta KE = 5.6 - 4

                         = 1.6J

This change in kinetic energy of the block = The total kinetic energy

       The workdone by the frictional force is

                  W_F = -F_f * d

the negative sign shown that the force is moving in the opposite direction

substituting 1.2 N for F_f  and 0.5 m for d

               W_F = -1.2*0.5

                     =-0.6J

Making Workdone by Force P the subject in the above equation we

            Workdone by Force P (W_P) = The total workdone(W_T)  - W_F

Substitution values

                    W_p = 1.6 -(-0.6)

                          W_p=2.2J

                   

                           

     

                 

4 0
3 years ago
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