or a bronze alloy, the stress at which plastic deformation begins is 267 MPa and the modulus of elasticity is 115 GPa. (a) What is the maximum load (in N) that may be applied to a specimen having a cross-sectional area of 164 mm2 without plastic deformation


Answer 1


The maximum load (in N) that may be applied to a specimen with this cross sectional area is F=43788 N


We know that the stress at which plastic deformation begins is 267 MPa.

We are going  to assume that the stress is homogeneously distributed. As a definition, we know that stress (P) is force (F) over area(A), as follows:

P=(F)/(A)    (equation 1)

We need to find the maximum load (or force) that may be applied before plastic deformation. And the problem says that the maximum stress before  plastic deformation is 267 MPa. So, we need to find the value of load when P= 267 MPa.

Before apply the equation, we need to convert the units of area in m^(2). So,

A[m^(2)]=A[mm^(2) ]((1 m)/(1000 mm)) ^(2)\nA[m^(2)]=164 mm^(2)((1 m)/(1000 mm))^(2) \nA[m^(2)]=0.000164 m^(2)

And then, from the equation 1,

F=PA\nF=(267 MPa)(0.000164 m^(2))\nF=(267x10^(6) Pa)(0.000164 m^(2))\nF=(267x10^(6) N/m^(2))(0.000164 m^(2))\nF=43788 N

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A case where debouncing can be compromised where a system is run partially through human intervention or that has different indications for one operation.

For example in a car wash management system, where green and red lights are used to indicate if a car is being washed, green light will be on and then red light means that there no car in washing que

Which statement best describes how power and work are related?O A. Power is the ability to do more work with less force.
O B. Power is a measure of how quickly work is done.
O C. Power and work have the same unit of measurement
O D. Power is the amount of work needed to overcome friction.
Pls answer quick



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4. Use voltage divider concepts to find the voltages indicated in the following circuits. You may want to use some of your results from problem 1. You may need to use the voltage divider equation more than once.


Hello, because there is not a circuit I'll explain the voltage divider and make an exercise, this way you can solve the problem using the method described here.

Answer with explanation:

A voltage divider uses the voltage distribution among components to find a voltage in a specific element of the circuit. If we have a source V1 connected to impedances Z1 and Z2 in series, we can use a voltage divider to find the voltage across Z1 or Z2 base on their value and the input voltage.

VZ1 = V1*Z1/(Z1+Z2)

VZ2 = V1*Z2/(Z1+Z2)

In the image, to find the voltage Vo across R2 we apply the following equation: Vo = (V1*R2)/(R1+R2).

To solve the exercise in the other image, we need to apply a voltage divider twice:

In-circuit 1 we are asked to find the voltage VAB that falls on R2 and R3 (the same voltage for both resistances because are in parallel), to do so we use a voltage divider using V1, R1 and RT where RT is the equivalent resistance RT = R2//R3 + R4, therefore, for circuit two VAC = (V1*R1)/(R1+RT). After finding VAC we apply voltage divider again to find VAB, see circuit 3, to do so we apply VAB = (VAC*R2//R3)/(R2//R3 + R4) = (VAC*R2//R3)/(RT)

The pressure in an automobile tire depends on thetemperature of the air in the tire. When the air temperature is25°C, the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m3, determine the pressure rise in the tire whenthe air temperature in the tire rises to 50°C. Also, determinethe amount of air that must be bled off to restore pressure toits original value at this temperature. Assume the atmosphericpressure to be 100 kPa.



The pressure rise in the tire when the air temperature in the tire rises to 50°C is 337.43 KPa.

The amount of air that must be bled off to restore pressure 0.007 Kg



T1 = 25°C = 298 K

T2 = 50°C = 323 K

volume of the tire = V = 0.025 m^(3)

P = 210 kPa (gage)

Pabs = 210 + 101 = 311 KPa

Before the temperature rise

P1 V1 = m1 R1 T1

m1 = (P1 V1)/(R1 T1) = (310 * 10^(3) * 0.25 )/(287 - 298) = 0.091 Kg\n \n

After the temperature rise

P2 = (m2 * R * T2)/(V2) = (0.091 *287*323 )/(0.025) = 337.43 KPa

after bleeding the pressure and the volume returns  to its first value

P1 = P2 and V1 = V2


(m2 * R * T2)/(V) = (m1 * R * T1)/(V)

m2 = (m1*T1)/(T2)

m2 = (0.091*298)/(332) = 0.084 Kg\n\n

mbleed = m1 - m2 --> mbleed = 0.91 - 0.84 = 0.007 Kg

P2 = 337.43 KPa

mbleed = 0.007 Kg

A cylindrical specimen of cold-worked steel has a Brinell hardness of 250.(a) Estimate its ductility in percent elongation.(b) If the specimen remained cylindrical during deformation and its original radius was 5 mm (0.20 in.), determine its radius after deformation.



A) Ductility = 11% EL

B) Radius after deformation = 4.27 mm


A) From equations in steel test,

Tensile Strength (Ts) = 3.45 x HB

Where HB is brinell hardness;

Thus, Ts = 3.45 x 250 = 862MPa

From image 1 attached below, for steel at Tensile strength of 862 MPa, %CW = 27%.

Also, from image 2,at CW of 27%,

Ductility is approximately, 11% EL

B) Now we know that formula for %CW is;

%CW = (Ao - Ad)/(Ao)

Where Ao is area with initial radius and Ad is area deformation.


%CW = [[π(ro)² - π(rd)²] /π(ro)²] x 100

%CW = [1 - (rd)²/(ro)²]

1 - (%CW/100) = (rd)²/(ro)²


(rd)²[1 - (%CW/100)] = (ro)²

So putting the values as gotten initially ;

(ro)² = 5²([1 - (27/100)]

(ro)² = 25 - 6.75

(ro) ² = 18.25

ro = √18.25

So ro = 4.27 mm

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?



The truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.


The distance that the truck starts slowing down = 80 ft from the stop sign

Using equations of motion, we can calculate the distance it will take the truck to stop, then check of it is less than or more than 80 ft.

u = initial velocity of the truck = 40 mph = 58.667 ft/s

v = final velocity of the truck = 0 ft/s (since it comes to a stop eventually)

x = horizontal distance covered during the deceleration

a = Deceleration = -12 ft/s² (it'll have a negative sign, since it is negative acceleration

v² = u² + 2ax

0² = 58.667² + 2(-12)(x)

24x = 3441.816889

x = 143.41 ft

143.41 ft > 80 ft; hence, the truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.

Corrected Question:

A truck traveling at 40 mph is approaching a stop sign. At time t₀ and at a distance of 80 ft, the truck begins to slow down by decelerating at 12 ft/s2, will the truck be able to stop in time?


The truck will not be able to stop in time.


==> First lets convert all variables to SI units

1 mph = 0.45m/s

40mph = 40 miles per hour = 40 x 0.45 m/s

40mph = 18m/s

1 ft = 0.3048m

80 ft = 80 x 0.3048m

80 ft = 24.38m


12ft/s² = 12 x 0.3048m/s²

12ft/s² = 3.66m/s²

==> Now, consider one of the equations of motion as follows;

v² = u² + 2as               -----------------(i)


v = final velocity of motion

u = initial velocity of motion

a= acceleration/deceleration of motion

s = distance covered during motion

Using this equation, lets calculate the distance, s, covered during the acceleration;

We know that;

v = 0               [since the truck comes to a stop]

u = 40mph = 18m/s

a = -12ft/s² = -3.66m/s²    [the negative sign shows that the truck decelerates]

Substitute these values into equation (i) as follows;

0² = 18² + 2 (-3.66)s

0 = 324 - 7.32s

7.32s = 324

s = (324)/(7.32)

s = 44.26m

The distance from where the truck starts decelerating to where it eventually stops is 44.26m which is past the stop sign (which is at 80ft = 24.38m).  This means that the truck stops, 44.26m - 24.38m = 19.88m, after the stop sign. Therefore, the truck will not be able to stop in time.