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kramer
3 years ago
10

In which situation will the friction between an object and the surface on which it is moving be decreased? a. the object is push

ed over a longer distance b. the weight of the moving object is decreased c. the object is pushed with greater force against the surface d. the surface on which the object is moving is made smoother Please put answers in the comments of actual answers, as I need them! Lots of points! 63 points
Physics
1 answer:
lawyer [7]3 years ago
5 0

d. the surface on which the object is moving is made smoother

This is called lubrication, I think ...

b. the weight of the moving object is decreased

If there is less araea of contact because of the weight reduction ("squeezing together") then this could also have an effexct. I thinl though the thrust of the q is on lubrication = oil, talcum powder, thin films of water etc

Please put answers in the comments of actual answers, as I need them! Lots of points! 63 points


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3 years ago
A monatomic ideal gas undergoes isothermal expansion from 0.08 m3 to 0.45 m3 at a constant temperature (initial pressure is 430
pochemuha

Answer :

(a) Internal energy change is, zero.

(b) Net heat transfer is, 59416.27 J

(c) Net work done is, -59416.27 J

Explanation : Given,

Initial volume of gas = 0.08m^3=80L

Final volume of the gas = 0.45m^3=45L

Initial pressure = 430 kPa

Moles of gas (n) = 1

First we have to calculate the temperature of the gas.

P_1V_1=nRT

where,

P_1 = initial pressure of gas

V_1 = initial volume of gas

n = moles of gas

R = gas constant = 8.314 kPa.L/K.mole

T = temperature of gas

Now put all the given values in the above formula, we get:

P_1V_1=nRT

(430kPa)\times (80L)=(1mole)\times (8.314kPa.L/K.mole)\times T

T=4137.59K

According to the question, this is the case of isothermal reversible expansion of gas.

As per first law of thermodynamic,

\Delta U=q+w

where,

\Delta U = internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

\Delta U=0

q=-w

The expression used for work done will be,

w=-nRT\ln (\frac{V_2}{V_1})

where,

w = work done on the system = ?

n = number of moles of gas  = 1

R = gas constant = 8.314 J/mole K

T = temperature of gas  = 4137.59 K

V_1 = initial volume of gas  = 0.08m^3

V_2 = final volume of gas  = 0.45m^3

Now put all the given values in the above formula, we get :

w=-1mole\times 8.314J/moleK\times 4137.59K\times \ln (\frac{0.45m^3}{0.08m^3})

w=-59416.27J

Net work done is, -59416.27 J

and,

q = - w = -(-59416.27 J) = 59416.27 J

Net heat transfer is, 59416.27 J

4 0
4 years ago
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