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Kapil S.

Tutor for Physics , Mathematics, Chemistry and Mechanical

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Basic Math

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Question:

If twice the son’s age in years is added to the father’s age, the sum is 70. But if the twice the father’s age is added to the son’s age, the sum is 95. Find the ages of father and son.

Kapil S.

Answer:

Strategy : Given word problem can be solved by converting it into a set of simultaneous equations. Solution : Let the age of son be X and that of the father be Y. then Equation 1: $$2X + Y = 70$$ by Condition 1 - twice the son’s age in years is added to the father’s age, the sum is 70 Equation 2: $$X + 2Y = 95$$ by Condition 2 - twice the father’s age is added to the son’s age, the sum is 95 Let's rearrange the equation 1 to get the value of Y in terms of X. $$ Y = 70 - 2X$$ Put this value of Y in Equation 2. $$X + 2 \times (70 - 2X) = 95$$ $$X + 140 - 4X = 95$$ $$140 - 3X = 95$$ $$140 - 95 = 3X$$ $$45=3X$$ $$X=15$$ ; Put this value of X in Equation 1 we get, $$2 \times 15 + Y = 70$$ $$ 30 + Y =70$$ $$ Y = 70-30$$ $$ Y = 40$$ Hence the age of son and father are 15 and 40 simultaneously.

Mechanical Engineering

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Question:

Air at 100 kPa, 290 K enters an ideal Otto cycle. The initial volume is 600 cm3. The compression ratio is 9.5, and the Temperature at the end of an isentropic expansion is 800 K. Find 1) Highest Temperature 2) Amount of heat added Use constant specific heat. Properties of air at room temperature: $$c_p = 1.005\times 10^3 J/kgK, c_v = 0.718 \times 10^3 J/kgK, k = 1.4, R = 0.287 \times 10^3 J/kgK. $$.

Kapil S.

Answer:

Given : Properties of air at room temperature: $$c_p = 1.005\times 10^3 J/kgK, c_v = 0.718 \times 10^3 J/kgK, k = 1.4, R = 0.287 \times 10^3 J/kgK. $$. Initial State :- $$P_1 = 100 kPa = 1 \times 10^5 N/m^2 , V_1 = 600 cm^3 = 6 \times 10^{-4} m^3 , T_1 = 290K. $$ Final State (end of expansion stroke) :- $$V_4=V_1 = 600 cm^3 = 6 \times 10^-4 m^3 , T_4 = 800K. $$ Compression Ratio$$(r)$$ :- 9.5 Strategy : 1) Highest Temperature in the Otto Cycle is at the state 3. The process from state 3 to 4 is isoentropic. Hence we can use the isoentropic ratios between these two states to find the $$T_3$$ highest temperature in the cycle. $$ \frac{T_3}{T_4} = \frac{V_4}{V_3} ^{k-1} = r ^{k-1}$$ , Here $$ k= 1.4 $$ and $$r=$$compression ratio. 2) For Otto Cycle heat addition takes place during State 2 to State 3. We can use First Law of Thermodynamics to calculate work done. $$ Q = M \times C_P \times \Delta T$$ Where $$C_p= 1.005\times 10^3 J/kgK$$ , $$M=$$Mass of air , $$\Delta T= T_3 - T_2$$. Mass $$M$$ can be found by ideal gas law $$ P \times V = M\times R \times T$$ for the state 1. Also, $$T_1$$ to $$T_2$$ is an isentropic process. Hence we can use the isoentropic ratios between these two states to find the $$T_2$$ $$ \frac{T_2}{T_1} = \frac{V_1}{V_2} ^{k-1} = r ^{k-1}$$ , Here $$ k= 1.4 $$ and $$r=$$compression ratio. Calculation: 1) $$ \frac{T_3}{T_4} = \frac{V_4}{V_3} ^{k-1} = r ^{k-1}$$\ $$ T_3 = T_4 \times r ^{k-1} = 800 \times 9.5 ^ {1.4 - 1}$$ $$ T_3 = 1968.7 K$$ Hence highest temperature is 1968.7 K. 2) $$ P_1 \times V_1 = M\times R \times T_1$$ $$ 1 \times 10^5 \times 6 \times 10^{-4} = M \times 0.287 \times 10^3 \times 290 $$ $$ M = 7.024 \times 10^{-4} kg$$ $$ \frac{T_2}{T_1} = \frac{V_1}{V_2} ^{k-1} = r ^{k-1}$$ $$ T_2 = T_1 \times r ^{k-1} = 290 \times 9.5 ^ {1.4 - 1}$$ $$ T_2 = 713.65 K$$ $$ Q = M \times C_P \times \Delta T$$ $$ Q = 7.024 \times 10^{-4} \times 1.005\times 10^3 \times ( 1968.7-713.65) $$ $$ Q = 0.650 \times 10^3 J = 0.650 \times kJ. $$

Physics

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Question:

A stationary block of mass 2 kg placed on a long frictionless horizontal table is pulled horizontally by a constant force F. It is found to move 10 m in the first two seconds. Find the magnitude of F.

Kapil S.

Answer:

This question belongs to Newton's Laws of motion and Kinematic Equations. Given : mass (M) = 2 Kg, distance moved (s) = 10 m and time (t) = 2 sec. To Find : The magnitude of force required (F) Strategy : We have to find the force (F) required to pull this block. As we know from Newtons Second law that $$ F = M \times a $$, where $$M$$ is mass and $$a$$ is acceleration. So in order to find force ($$F$$) we need mass ($$M$$) and acceleration ($$a$$) of the block. We already know the mass ($$M$$) of the block. Hence all we have to do is find acceleration ($$a$$) of the block using distance ($$s$$) moved by the block in time ($$t$$). From second equation of motion $$s = u\times t + 1/2 \times a\times t^2$$, where u is the initial velocity. As given in the problem, initially the block is stationary which implies $$u= 0 $$. Hence we can find $$a$$ using this equation and put it back into $$ F = M \times a $$ to get the required force. Calculation : $$s = u\times t + 1/2 \times a\times t^2$$ $$10 = 0\times 2 + 1/2 \times a\times 2^2$$ $$10 = 2 \times a$$ $$a = 5 m/sec^2 $$ $$ F = M \times a $$ $$ F = 2 \times 5 $$ $$ F = 10 N$$. Hence the force of 10N will be required to pull the block.

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