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2 years ago

A steel ball of mass m falls from a height h

Physics

1 answer:

SOVA2 [1]2 years ago

7 0

Answer:

The average force F_avg = m*g

Explanation:

Given:

- Mass of the ball = m

- Initially at height = h

- Dropped from rest v_i = 0

Find:

The average force F_avg that the ball imparts to the scale on every hit.

Solution:

- The velocity of the ball just before it hits the scale can be evaluated by the energy conservation as follows:

E.K_1 + E.P_1 = E.K_2 + E.P_2

- Where, E.K = 0 ..... initial velocity V_i = 0

E.P = 0 ...... Scale is the datum reference.

Hence,

E.P_1 = E.K_2

m*g*h = 0.5*m*V_o^2

V_o = sqrt ( 2*g*h)

- We will apply the conservation of momentum on the ball just before it hits the scale and just after produces an impulse. The impulse is given by Newton's second law of motion:

Ft = m*(V_f - V_o )

Where, Ft is the impulse. V_f velocity after impact. V_o = V_f (no loss of energy upon impact) - assumption.

- Taking upward direction as +.

Ft = m*(V_o + V_o ) = 2*m*V_o

Ft = 2*m*sqrt ( 2*g*h)

- The total time between two successive hits T can be calculated by using second kinematic equation of motion in vertical direction, as follows:

y = y(0) + V_o*T - 0.5*g*T^2

Where, y = y(0) = 0 ... start and end points same.. Datum

Hence,

0 = 0+ V_o*T - 0.5*g*T^2

0 = V_o - 0.5*g*T

T = 2*V_o / g = 2*sqrt ( 2*g*h) / g

- The average force F_avg can now be calculated using the expression given:

F_avg*T = Ft

F_avg = Ft / T

F_avg = [2*m*sqrt ( 2*g*h)] / [2*sqrt ( 2*g*h) / g]

F_avg = m*g

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2 years ago

200-grams of computer chips with a specific heat of 0.3 kJ/kg·K are initially at 25°C. These chips are cooled by placement in 0.

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a. -0.01324 kJ/K, b. = 0.03233 kJ/K , c. = 0.01909, Yes the process is possible

Explanation:

Heat transfer will occur between the chip and the surrounding fluid. Then, finally they will attain a common equilibrium temperature and heat transfer will stop. Now, if we assume that, after heat transfer, chip will attain the temperature of fluid, that is, -34 C,, So , to check whether this is possible

Amount of energy lost by the chip = m . c . (T(i) - T(f))

= 0.2 x 0.3 (25 + 34) = 3.54 KJ

Now, to evaluate the final state of the fluid, after the heat transfer completion,

Energy Gained = m(mew final – mew initial) = m[(μf+ x . μfg) - μf]

Note that heat transfer will change the internal energy of the fluid. Do not consider enthalpy change, as this is not a problem involving fluid flow in and out of the system

M[(μf+ x . μfg) - μf] = m(xμfg)

Energy gained by the fluid will be equal to the energy lost by the chip (No energy loss to the surroundings)

3.54 = 0.1 . X x 203.29

x = 0.1741, which is the dryness fraction of fluid at the final state.

Observe that the total energy lost by the chips is 3.45 kJ and fluid R-134a has got its value of mew fg at -34 C which is = 203.29 kJ/kg

So for 0.1kg of R-134a

0.1 x μfg= 20.329 kJ, which is much greater than 3.45 kJ, therefore, it is certain that the state of fluid will be at -34 C only and at the saturation pressure of 69.56 KPa. So the chip will come to attain the temperature of -34 C.

a. Write the equation for the change of entropy in the chips

ΔSchips = mchips . c . ln(T2/T1), where mc is the mass of chips, c is the specific heat of chips, T2 is the temperature at state 2 and T1 is the temperature at state 1

Substitute mc = 0.2 kg, c = 0.3kJ/kg.K, T1 = 25 + 273, T2 = -34 + 273

delSchips = 0.2 x 0.3 x ln [(-34+273)/ (25+273)]

= -0.01324 kJ/K

There fore the change in entropy of the chips is -0.01324 kJ/K

b. Entropy change of fluid R- 134a

ΔS2 = m[Sfinal – S initial]

= m[Sf + x . Sfg - Sf]

= 0.2 x (0.1741 x 0.92859)

= 0.03233 kJ/K

c. Calculate the total change in the entropy of the entire system

delS = delSchips + delSR -134a

= -0.01324 + 0.03233

= 0.01909

Since the total change in entropy of the entire system is positive that exactly explains that the actual processes are happening in the direction of increase of entropy therefore, the process is possible.

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