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3 years ago

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The sports car has a weight of 4900 lblb and center of gravity at GG. If it starts from rest it causes the rear wheels to slip a

s it accelerates. The coefficients of static and kinetic friction at the road are μsμs = 0.5 and μkμkmu_k = 0.35, respectively. Neglect the mass of the wheelDetermine how long it takes for it to reach a speed of 10 ft/s.What are the normal reactions at each of the rear wheels on the road?What are the normal reactions at each of the front wheels on the road?

1 answer:

Zinaida [17]3 years ago

3 0

Answer:

2.27secs

Explanation:

The car will take 2.27secs to reach a speed of 10 ft/s.

Kindly go through the attached files for a step by step solution that will show you how the answer was gotten and the normal reaction of both the front and rear wheels.

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Carbon dioxide (CO2) is compressed in a piston-cylinder assembly from p1 = 0.7 bar, T1 = 320 K to p2 = 11 bar. The initial volum

tekilochka [14]

Answer:

$W_{12}=-53.9056KJ$

Part A:

$Q=-7.03734 KJ/Kg$ (-ve sign shows heat is getting out)

Part B:

$Q=1.5265KJ/Kg$ (Heat getting in)

The value of Q at constant specific heat is approximately 361% in difference with variable specific heat and at constant specific heat Q has opposite direction (going in) than Q which is calculated in Part B from table A-23. So taking constant specific heat is not a good idea and is questionable.

Explanation:

Assumptions:

Gas is ideal
System is closed system.
K.E and P.E is neglected
Process is polytropic

Since Process is polytropic so $W_{12} =\frac{P_{2}V_{2}-P_{1}V_{1}}{1-n}$

Where n=1.25

Since Process is polytropic :

$\frac{V_{2}}{V_{1}}=(\frac{P_{1}}{P_{2}})^{\frac{1}{1.25}} \\V_{2}= (\frac{P_{1}}{P_{2}})^{\frac{1}{1.25}} *V_{1}$

$V_{2}= (\frac{0.7}{11})^{\frac{1}{1.25}} *0.262\\V_{2}=0.028924 m^3$

Now, $W_{12} =\frac{P_{2}V_{2}-P_{1}V_{1}}{1-n}$

$W_{12} =\frac{11*0.028924-0.7*0.262}{1-1.25}(\frac{10^{5}N/m^2}{1 bar})(\frac{1 KJ}{10^{3}Nm})$

$W_{12}=-53.9056KJ$

We will now calculate mass (m) and Temperature T_2.

$m=\frac{P_{1}V_{1}}{RT_{1}}\\ m=\frac{0.7*0.262}{\frac{8.314KJ}{44.01Kg.K}*320}(\frac{10^{5}N/m^2}{1 bar})(\frac{1 KJ}{10^{3}Nm})\\m=0.30338Kg$

$T_{2} =\frac{P_{2}V_{2}}{Rm}\\ m=\frac{11*0.028924}{\frac{8.314KJ}{44.01Kg.K}*0.30338}(\frac{10^{5}N/m^2}{1 bar})(\frac{1 KJ}{10^{3}Nm})\\T_{2} =555.14K$

Part A:

According to energy balance::

$Q=mc_{v}(T_{2}-T_{1})+W_{12}$

From A-20, C_v for Carbon dioxide at 300 K is 0.657 KJ/Kg.k

$Q=0.30338*0.657(555.14-320)+(-53.9056)$

$Q=-7.03734 KJ/Kg$ (-ve sign shows heat is getting out)

Part B:

From Table A-23:

$u_{1} at 320K = 7526 KJ/Kg$

$u_{2} at 555.14K = 15567.292$ (By interpolation)

$Q=m(\frac{u(T_{2})-u(T_{1})}{M} )+W_{12}$

$Q=0.30338(\frac{15567.292-7526}{44.01} )+(-53.9056)$

$Q=1.5265KJ/Kg$ (Heat getting in)

The value of Q at constant specific heat is approximately 361% in difference with variable specific heat and at constant specific heat Q has opposite direction (going in) than Q which is calculated in Part B from table A-23. So taking constant specific heat is not a good idea and is questionable.

7 0

4 years ago

If a resistor only has three color bands, it has a tolerance of 20 percent. What would this mean for a resistor specified at 10,

Norma-Jean [14]

B is the correct answe it will fail if more than 12.000
Hope this helps
:)

3 0

3 years ago

Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stre

velikii [3]

Complete Question

Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a [111] direction, compute the stress at which the crystal yields if its critical resolved shear stress is 2.9 MPa.

Answer:

The stress is $\sigma = 10. 655 MPa$

Explanation:

From the question we are told that

The critical yield resolved shear stress is $\sigma = 2.9Mpa$

First we obtain the angle $\lambda$ between the slip direction [121] and [111]

$\lambda = cos^{-1} [\frac{(u_1 u_2 + v_1 v_2 + w_1 w_2}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_2^2 + v_2^2 + w_2 ^2)} } ]$

Where $u_1 ,u_2 ,v_1 , v_2 , w_1 , w_2$ are the directional indices

$\lambda = cos ^-[ \frac{(1) (-1) + (2) (1) + (1) (1)}{\sqrt{((1)^2 +(2)^2 + (1)^2)}\sqrt{((-1)^2 + (1)^2 + (1)^2 ) } } ]$

$= cos^{-1} [\frac{2}{\sqrt{6} \sqrt{3} } ]$

$= 61.87^0$

Next is to obtain the angle $\O$ between the direction [121] and [101]

$\O = cos^{-1} [\frac{(u_1 u_3 + v_1 v_3 + w_1 w_3}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_3^2 + v_3^2 + w_3 ^2)} } ]$

Substituting 1 for $u_1$ , 2 for $v_1$ , 1 for $w_1$ , 1 for $u_2$ , 0 for $v_2$ , and 1 for $w_2$

$\O = cos^{-1} [\frac{1* 1 + 2*0 + 1*1 }{\sqrt{1^2 + 2^2 + 1^2 } \sqrt{(1^2 + 0^2 + 1^2 )} } ]$

$\O = cos^{-1} [\frac{2}{\sqrt{6} \sqrt{2} } ]$

$= 54.74 ^o$

The stress is mathematically represented as

$\sigma = \frac{\tau_c}{cos \O cos \lambda }$

$= \frac{2.9}{cos 54.74^o cos 61.87^o}$

$= \frac{2.9}{0.2722}$

$\sigma = 10. 655 MPa$

5 0

3 years ago

Nitrogen (N2) contained in a piston–cylinder arrangement, initially at 10 bar and 405 K, undergoes an expansion to a final tempe

yKpoI14uk [10]

Answer:28.21 kJ/kg

Explanation:

Given

$P_1=10\ bar$

$T_1=405\ K$

$T_2=300\ K$

Process $PV^{1.3}=constant$

Work done for Polytropic process

$W=\dfrac{P_1V_1-P_2V_2}{n-1}$

where n=Polytropic index

$W=\dfrac{R(T_1-T_2)}{n-1}$

$W=\dfrac{0.296(405-300)}{1.3-1}\quad [R_{N_2}=\frac{8.314}{28}]$

$W=103.6\ kJ\kg$

Now Calculating change in Internal energy

$\Delta U=c_v(T_2-T_1)$

$\Delta U=0.718\times (300-405)$

$\Delta U=-75.39\ kJ/kg$

Now applying First law concept

$\Delta U=Q-W$

$Q=W+\Delta U$

$Q=103.6-75.392$

$Q=28.21\ kJ/kg$

6 0

4 years ago

Select the four parts of the plasma system that

Mandarinka [93]

Answer:

Power

pressure

cup

Temperature

Explanation:

6 0

3 years ago