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2 years ago

What is the net force acting on a car cruising at a constant velocity of 70 km/h (a) on a level road and (b) on an uphill road?

Engineering

1 answer:

ElenaW [278]2 years ago

4 0

Answer:

a) zero b) zero

Explanation:

Newton's first law tells us that a body remains at rest or in uniform rectilinear motion, if a net force is not applied on it, that is, if there are no applied forces or If the sum of forces acting is zero. In this case there is a body that moves with uniform rectilinear motion which implies that there is no net force.

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A rigid tank contains an ideal gas at 40°C that is being stirred by a paddle wheel. The paddle wheel does 240 kJ of work on the

Sergeu [11.5K]

To solve the problem it is necessary to consider the concepts and formulas related to the change of ideal gas entropy.

By definition the entropy change would be defined as

$\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{P_2}{P_1})$

Using the Boyle equation we have

$\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{v_1T_2}{v_2T_1})$

Where,

$C_p$ = Specific heat at constant pressure

$T_1$ = Initial temperature of gas

$T_2$ = Final temprature of gas

R = Universal gas constant

$v_1$ = Initial specific Volume of gas

$v_2$ = Final specific volume of gas

According to the statement, it is an isothermal process and the tank is therefore rigid

$T_1 = T_2, v_2=v_1$

The equation would turn out as

$\Delta S = C_p ln1-ln1$

<em>Therefore the entropy change of the ideal gas is 0</em>

Into the surroundings we have that

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat Exchange

T = Temperature in the surrounding

Replacing with our values we have that

$\Delta S = \frac{230kJ}{(30+273)K}$

$\Delta S = 0.76 kJ/K$

<em>Therefore the increase of entropy into the surroundings is 0.76kJ/K</em>

8 0

2 years ago

Only put coolant into your radiator when the engine is _____.

Arada [10]

Only put ciilant into ur radiator when the engine is cool (D)

5 0

3 years ago

Windmills slow the air and cause it to fill a larger channel as it passes through the blades. Consider a circular windmill with

Scilla [17]

Answer:

DIAMETER = 9.797 m

POWER = $\dot W = 28.6 kW$

Explanation:

Given data:

circular windmill diamter D1 = 8m

v1 = 12 m/s

wind speed = 8 m/s

we know that specific volume is given as

$v =\frac{RT}{P}$

where v is specific volume of air

considering air pressure is 100 kPa and temperature 20 degree celcius

$v = \frac{0.287\times 293}{100}$

v = 0.8409 m^3/ kg

from continuity equation

$A_1 V_1 = A_2 V_2$

$\frac{\pi}{4}D_1^2 V_1 = \frac{\pi}{4}D_1^2 V_2$

$D_2 = D_1 \sqrt{\frac{V_1}{V_2}}$

$D_2 = 8 \times \sqrt{\frac{12}{8}}$

$D_2 = 9.797 m$

mass flow rate is given as

$\dot m = \frac{A_1 V_1}{v} = \frac{\pi 8^2\times 12}{4\times 0.8049}$

$\dot m = 717.309 kg/s$

the power produced $\dot W = \dot m \frac{ V_1^2 - V_2^2}{2} = 717.3009 [\frac{12^2 - 8^2}{2} \times \frac{1 kJ/kg}{1000 m^2/s^2}]$

$\dot W = 28.6 kW$

8 0

3 years ago

A DC generator turns at 2000 rpm and has an output of 200 V. The armature constant is 0.5 V-min/Wb, and the field constant of th

WITCHER [35]

Answer:

b. 10A

Explanation:

Using the formula, E= k × r×I

200= 0.5 ×2000×0.02×I

200=20×I

Dividing with 20

I = 200/20= 10A

4 0

3 years ago

Read 2 more answers

Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging wat

zlopas [31]

Answer: $10.631\times 10^3\ N/m^2$

Explanation:

Given

Discharge is $Q=12.5\ L$

Diameter of pipe $d=150\ mm$

Distance between two ends of pipe $L=800\ m$

friction factor $f=0.008$

Average velocity is given by

$\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s$

Pressure difference is given by

$\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow \Delta P=10.631\ kPa$

8 0

2 years ago