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3 years ago

5

To drive your car from Auburn to Atlanta (173.8 km) takes a significant amount of work. Assuming the force of friction between t

he tires and the road is 10290 N, how much work needed to drive that far? Express your answer in megajoules (MJ).

1 answer:

Daniel [21]3 years ago

5 0

Answer:

1788.402 MJ

Explanation:

Work done = Force (N) x distance (m)

First we have to convert distance into metres:

173.8 x 1000 = 173,800 m

Then plug these values into the equation above:

173,800 x 10290 = 1788402000 J

The reason it's Joules (the unit for energy) is because work done = energy transferred

Now we have to convert Joules into Mega Joules:

1 J = 1/1000000 MJ

1788402000 J = 1788402000/ 1000000 = 1788.402 MJ

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3 years ago

What is the concentration of H^ + at apH = 2 ? Mol / L What is the concentration of H^ + ions at apH = 6 ? Mol/L How many more H

Nonamiya [84]

Answer:

The concentration of hydrogen ion at pH is equal to 2 : $= [H^+]=0.01 mol/L$

The concentration of hydrogen ion at pH is equal to 6 : $[H^+]'=0.000001 mol/L$

There are 0.009999 more moles of $H^+$ ions in a solution at a pH = 2 than in a solution at a pH = 6.

Explanation:

The pH of the solution is the negative logarithm of hydrogen ion concentration in an aqueous solution.

$pH=-\log [H^+]$

The hydrogen ion concentration at pH is equal to 2 = [H^+]

$2=-\log [H^+]\\$

$[H^+]=10^{-2}M= 0.01 M=0.01 mol/L$

The hydrogen ion concentration at pH is equal to 6 = [H^+]

$6=-\log [H^+]\\\\$

$[H^+]=10^{-6}M= 0.000001 M= 0.000001 mol/L$

Concentration of hydrogen ion at pH is equal to 2 = $[H^+]=0.01 mol/L$

Concentration of hydrogen ion at pH is equal to 6 = $[H^+]'=0.000001 mol/L$

The difference between hydrogen ion concentration at pH 2 and pH 6 :

$= [H^+]-[H^+]' = 0.01 mol/L- 0.000001 mol/L = 0.009999 mol/L$

Moles of hydrogen ion in 0.009999 mol/L solution :

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There are 0.009999 more moles of $H^+$ ions in a solution at a pH = 2 than in a solution at a pH = 6.

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3 years ago

A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com

bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature $=20^{\circ}C$

Final Temperature $=80^{\circ}C$

mass flow rate of cold fluid $\dot{m_c}=1.2 kg/s$

Initial Geothermal water temperature $T_h_i=160^{\circ}C$

Let final Temperature be T

mass flow rate of geothermal water $\dot{m_h}=2 kg/s$

diameter of inner wall $d_i=1.5 cm$

$U_{overall}=640 W/m^2K$

specific heat of water $c=4.18 kJ/kg-K$

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

$\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)$

$2\times (160-T)=1.2\times (80-20)$

$160-T=36$

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As heat exchanger is counter flow therefore

$\Delta T_1=160-80=80^{\circ}C$

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$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}$

$LMTD=\frac{80-104}{\ln \frac{80}{104}}$

$LMTD=91.49^{\circ}C$

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

$\dot{m_c}c(80-20)=U\cdot A\cdot$ (LMTD)

$A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2$

$A=\pi DL=5.144$

$L=\frac{5.144}{\pi \times 0.015}$

$L=109.16 m$

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