ENGINEERING COLLEGE

Calculate the angle of banking on a bend of 100m radius so that vehicles can travel round the bend at 50km/hr without side thrust on the tyres.

Answers

Answer 1

Answer:

Answer:

11.125°

Explanation:

Given:

Radius of bend, R = 100 m

Speed around the bend = 50 Km/hr = $(5)/(18)*50$ = 13.89 m/s

Now,

We have the relation

$\tan\theta=(v^2)/(gR)$

where,

θ = angle of banking

g is the acceleration due to gravity

on substituting the respective values, we get

$\tan\theta=(13.89^2)/(9.81*100)$

$\tan\theta=0.1966$

θ = 11.125°

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Q1. A truck traveling at 40 mph is approaching a stop sign. At time ????0 and at a distance of 80ft, the truck begins to slow down by decelerating rate of 12 ft/sec2 . Will the truck be able to stop in time?
Shear modulus is analogous to what material property that is determined in tensile testing? (a)- Percent reduction of area (b) Yield strength (c)- Elastic modulus (d)- Poisson's ratio
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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.

Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 39 m3/min and exits at 12 bar, 400 K. Heat transfer occurs at a rate of 6.5 kW from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in kW.

Answers

Answer:

The power input, in kW is -86.396 kW

Explanation:

Given;

initial pressure, P₁ = 1.05 bar

final pressure, P₂ = 12 bar

initial temperature, T₁ = 300 K

final temperature, T₂ = 400 K

Heat transfer, Q = 6.5 kW

volumetric flow rate, V = 39 m³/min = 0.65 m³/s

mass of air, m = 28.97 kg/mol

gas constant, R = 8.314 kJ/mol.k

R' = R/m

R' = 8.314 /28.97 = 0.28699 kJ/kg.K

Step 1:

Determine the specific volume:

p₁v₁ = RT₁

$v_1 = (R'T_1)/(p_1) = ((0.28699.(kJ)/(kg.K) )(300 k))/((1.05 bar *\ (10^5 N/m^2)/(1 bar) *(1kJ)/(1000N.m) )) \n\nv_1 = 0.81997 \ m^3/kg$

Step 2:

determine the mass flow rate; m' = V / v₁

mass flow rate, m' = 0.65 / 0.81997

mass flow rate, m' = 0.7927 kg/s

Step 3:

using steam table, we determine enthalpy change;

h₁ at T₁ = 300.19 kJ /kg

h₂ at T₂ = 400.98 kJ/kg

Δh = h₂ - h₁

Δh = 400.98 - 300.19

Δh = 100.79 kJ/kg

step 4:

determine work input;

W = Q - mΔh

Where;

Q is heat transfer = - 6.5 kW, because heat is lost to surrounding

W = (-6.5) - (0.7927 x 100.79)

W = -6.5 -79.896

W = -86.396 kW

Therefore, the power input, in kW is -86.396 kW

You are analyzing an open-return wind tunnel that intakes air at 20 m/s and 320K. When the flow exits the wind tunnel it is moving at a speed of 250 m/s. What is the temperature of the flow exiting that wind tunnel

Answers

The solution is in the attachment

Answer:

please find attached.

Explanation:

An inflatable structure has the shape of a half-circular cylinder with hemispherical ends. The structure has a radius of 40 ft when inflated to a pressure of 0.60 psi. A longitudinal seam runs the entire length of the structure. The seam fails in tension when the load is 600 pounds per inch of seam. What is the factor of safety with respect to longitudinal seam failure?

Answers

Find the given attachment

8.2.1: Function pass by reference: Transforming coordinates. Define a function CoordTransform() that transforms the function's first two input parameters xVal and yVal into two output parameters xValNew and yValNew. The function returns void. The transformation is new

Answers

Answer:

The output will be (3, 4) becomes (8, 10)

Explanation:

#include <stdio.h>

//If you send a pointer to a int, you are allowing the contents of that int to change.

void CoordTransform(int xVal,int yVal,int* xNew,int* yNew){

*xNew = (xVal+1)*2;

*yNew = (yVal+1)*2;

}

int main(void) {

int xValNew = 0;

int yValNew = 0;

CoordTransform(3, 4, &xValNew, &yValNew);

printf("(3, 4) becomes (%d, %d)\n", xValNew, yValNew);

return 0;

}

An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a lagging power factor of 0.77. Determine the size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging.

Answers

Answer:

$Q=41.33 KVAR\ \nat\n\ 480 Vrms$

Explanation:

From the question we are told that:

Voltage $V=480/0 \textdegree V$

Power $P=120kW$

Initial Power factor $p.f_1=0.77 lagging$

Final Power factor $p.f_2=0.9 lagging$

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

$p.f_1=0.77$

$cos \theta_1 =0.77$

$\theta_1=cos^(-1)0.77$

$\theta_1=39.65 \textdegree$

And

$p.f_2=0.9$

$cos \theta_2 =0.9$

$\theta_2=cos^(-1)0.9$

$\theta_2=25.84 \textdegree$

Therefore

$Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=-41.33VAR$

Therefore

The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

$Q=41.33 KVAR\ \nat\n\ 480 Vrms$

Describe how a feeler gauge can be used to assist in the adjustment of a spark plug electrode gap

Answers

Answer:

Explanation:

Adjusting the distance between the two electrodes is called gapping your spark plugs. You need a feeler gauge to gap your spark plugs properly If you're re-gapping a used plug, make sure that it's clean (gently scrub it with a wire brush)

hope this you

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