ENGINEERING COLLEGE

The sports car has a weight of 4500-lb and a center of gravity at G. If it starts from rest it causes the rear wheels to slip as it accelerates. Determine how long it takes for it to reach a speed of 10 ft/s. Also, what are the normal reactions at each of the four wheels on the road? The coefficients of static and kinetic friction at the road are us=0.5 and uk=0.3, respectively. Neglect the mass of the wheels.

Answers

Answer 1

Answer:

Answer:

Time=2.72 seconds

Front wheel reactions= 1393 lb

Rear wheel reactions= 857 lb

Explanation:

The free body diagram is assumed to be the one attached here

The mass, m of the car is

$M=\frac {W}{g}$ where W is weight and g is acceleration due to gravity

Taking g as $32.2 ft/s^(2)$ then

$M=\frac {4500}{32.2}=139.75 lbm$

Considering equilibrium in x-axis

$Ma_G-f=0$

$Ma_G-(\mu_g* 2N_B)=0$

$139.75* a_G-(0.3* 2* N_B)=0$

$0.6N_B=139.75a_g$

$N_B=232.92a_g$

At point A using the law of equilibrium, the sum of moments is 0 hence

$-2N_B(6)+4500(2)=-Ma_G(2.5)$

$-12N_B+9000=-139.75a_G* 2.5$

$-12(232.92a_G)+900=-349.375a_G$

$a_g\approx 3.68 ft/s^(2)$

The normal reaction at B is therefore

$N_B=232.92a_G=232.92* 3.68\approx 857 lb$

Consider equilibrium in y-axis

$4500-2N_A-2N_B=0$

$N_A+N_B=2250$

$N_A+857=2250$

$N_A=1393 lb$

To find time that the car takes to a speed of 10 ft/s

Using kinematic equation

V=u+at

10=0+3.68t

$t=\frac {10}{3.68}\approx 2.72 s$

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How much computer memory (in bytes) in minimum would be required to store 10 seconds of a sensor signal sampled by a 12-bit A/D converter operating at a sampling rate of 5 kHz?

Answers

Answer:

73.24 K byte

Explanation:

Assuming that

N = total number of samples

N = 10 * 5kHz

N = 50*10^3

Also, the total number of bits, T

T = 12 * N

T = 12 * 50*10^3

T = 600 * 10^3

And then, finally, the total number of byte,

B = 600*10^(3/8)

B = 75*10^3 byte

75*10^3 byte = 75*10^3/1024 kilo byte

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A circuit contains a 40 ohm resistor and a 60 ohms resistor connected in parallel. If you test this circuit with a DMM you should read a total resistance

Answers

Answer:

24Ω

Explanation:

When resistors are connected in parallel, the reciprocal of their combined resistance, when read with a DMM (Digital Multimeter - for measuring various properties of a circuit or circuit element such as resistance...) is the sum of the reciprocals of their individual resistances.

For example if two resistors of resistances R₁ and R₂ are connected together in parallel, the reciprocal of their combined resistance Rₓ is given by;

$(1)/(R_x)$ = $(1)/(R_1)$ + $(1)/(R_2)$

Solving for Rₓ gives;

$R_(x)$ = $(R_1 * R_2)/(R_1 + R_2)$ ------------------(i)

From the question;

Let

R₁ = resistance of first resistor = 40Ω

R₂ = resistance of second resistor = 60Ω

Now,

To get their combined or total resistance, Rₓ, substitute these values into equation (i) as follows;

$R_(x)$ = $(40 * 60)/(40 + 60)$

$R_(x)$ = $(2400)/(100)$

$R_(x)$ = 24 Ω

Therefore, the total resistance is 24Ω

Answer:

24 ohms

Explanation:

hope this helped :)

God bless you!!!!

Water needs to be turned into steam in a high altitude lab where the atmospheric pressure is 84.6 KPa. Computte the heat energy (in calories) required to evaporate 900g of water at 15 degree C under these conditions.

Answers

Answer:

558.1918 kilocalories = 558191.8 calories

Explanation:

Data provided in the question:

Atmospheric pressure = 84.6 KPa

Mass of water, m = 900 g = 0.90 kg

Temperature = 15°C

Now,

Temperature at 84.6 KPa = 94.77°C

Therefore,

Heat energy required = m(CΔT + L)

here,

C is the specific heat of the water = 4.2 KJ/kg.°C

L = Latent heat of water = 2260 KJ/kg

Thus,

Heat energy required = 0.90[ 4.2 × (94.77 - 15) + 2260 ]

= 2335.53 KJ

also,

1 KJ = 0.239 Kilocalories

Therefore,

2335.53 KJ = 0.239 × 2335.53 Kilocalories

= 558.1918 kilocalories = 558191.8 calories

At full load, a commercially available 100hp, three phase induction motor operates at an efficiency of 97% and a power factor of 0.88 lag. The motor is supplied from a three-phase outlet with a line voltage rating of 208V.a. What is the magnitude of the line current drawn from the 208 V outlet? (1 hp = 746 W.) b. Calculate the reactive power supplied to the motor.

Answers

Answer:

I = Line Current = 242.58 A

Q = Reactive Power = 41.5 kVAr

Explanation:

Firstly, converting 100 hp to kW.

Since, 1 hp = 0.746 kW,

100 hp = 0.746 kW x 100

100 hp = 74.6 kW

The power of a three phase induction motor can be given as:

$P_(in) = √(3) VI Cos\alpha\n$

where,

P in = Input Power required by the motor

V = Line Voltage

I = Line Current

Cosα = Power Factor

Now, calculating Pin:

$efficiency = \frac{{P_(out)} }{P_(in) }\n0.97 = (74.6)/(P_(in) ) \nP_(in) = (74.6)/(0.97)\n P_(in) = 76.9 kW$

a) Calculating the line current:

$P_(in) = √(3)VICos\alpha \n76.9 * 1000= √(3)*208*I*0.88\nI = (76.9*1000)/(√(3)*208*0.88 )\nI = 242.58 A$

b) Calculating Reactive Power:

The reactive power can be calculated as:

Q = P tanα

where,

Q = Reactive power

P = Active Power

α = power factor angle

Since,

$Cos\alpha =0.88\n\alpha =Cos^(-1)(0.88)\n\alpha=28.36$

Therefore,

$Q = 76.9 * tan (28.36)\nQ = 76.9 * (0.5397)\nQ = 41. 5 kVAr$

Consider a mixture of hydrocarbons that consists of 60 percent (by volume) methane, 30 percent ethane, and 10 percent propane. After passing through a separator, the mole fraction of the propane is reduced to 1 percent. the mixture pressure before and after the separation is 100 kPa. Determine the change in the partial pressures of all the constituents in the mixture.

Answers

Answer:

$\Delta P_m=6\text{kPa}\n\Delta P_e=3\text{kPa}\n\Delta P_p=-9\text{kPa}$

Explanation:

mole fraction of propane after passing through the separator is $\beta_p$

$(\beta)/(0.6+0.3+\beta)=0.01$

$\beta =9.09* 10^-^3$

mole fractions of ethane $\beta _e$ and methane $\beta_m$ after passing through separator are:

$\beta_e =(0.3)/(0.3+0.6+0.00909)=0.66\n\beta_m=(0.6)/(0.3+0.6+0.00909)=0.33$

Change in partial pressures then can be written as:

$\Delta P=(y_2-y_1)\cdot P$ where $y_2$ and $y_1$ are mole fractions after and before passing through the separator

Hence,

$\Delta P_m=(0.66-0.6)\cdot 100\text{k}=6\text{kPa}\n\Delta P_e=(0.33-0.3)\cdot 100\text{k}=3\text{kPa}\n\Delta P_p=(0.01-0.1)\cdot 100\text{k}=-9\text{kPa}$

. Were you able to observe ???? = 0 in the circuit you constructed during lab? Why or why not? Hint: What value of resistance would be needed for ???? = 0? 2. What feature in the time response of an RLC circuit distinguishes a critically damped response from an underdamped response? 3. Why must an op-amp be powered to be used in a circuit? 4. If you were handed a parts kit with an unknown op-amp, what information would you need to find prior to using it in a circuit?

Answers

Answer:

an attachment is below

Explanation:

1) the formula for damping coefficient id for RLC series circuit.

For \xi =0 you can make c=0 but inductor will still have some capacitance.

2) the responses of critically damped system and under damped system are shown with comments on their time response.

4) There can be many different answers to this question, but the 4 I have mentioned are the most important parameters we need to know about an unknown op-amp if we are to use it in our circuit.

Hope it answers all your questions.

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