Answer:

**Answer:**

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

**Explanation:**

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:

(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 . So,

And then, from the equation 1,

1.19. A gas is confined in a 0.47 m diameter cylinder by a piston, on which rests a weight. The mass of the piston and weight together is 150 kg. The local acceleration of gravity is 9.813 m·s−2, and atmospheric pressure is 101.57 kPa. (a) What is the force in newtons exerted on the gas by the atmosphere, the piston, and the weight, assuming no friction between the piston and cylinder? (b) What is the pressure of the gas in kPa? (c) If the gas in the cylinder is heated, it expands, pushing the piston and weight upward. If the piston and weight are raised 0.83 m, what is the work done by the gas in kJ? What is the change in potential energy of the piston and weight?\

A cylindrical bar of metal having a diameter of 19.2 mm and a length of 207 mm is deformed elastically in tension with a force of 52900 N. Given that the elastic modulus and Poisson's ratio of the metal are 61.4 GPa and 0.34, respectively, determine the following: a. The amount by which this specimen will elongate in the direction of the applied stress. b. The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.

Reduce the following lambda-calculus term to the normalform. Show all intermediate steps, with one beta reduction at a time. In the reduction, assume that you are supplied with extra rules thatallow you to reduce the multiplication of two natural numbers into thecorresponding result.(λf.λx.f(f x))(λy.y≠3) 2

A hydraulic jump is induced in an 80 ft wide channel. The water depths on either side of the jump are 1 ft and 10 ft. Please calculate: a) The velocity of the faster moving flow. b) The flow rate (discharge). c) The Froude number of the sub-critical flow. d) The flow energy dissipated in the hydraulic jump (expressed as percentage of the energy prior to the jump). e) The critical depth.

A site is compacted in the field, and the dry unit weight of the compacted soil (in the field) is determined to be 18 kN/m3. Determine the relative compaction if the maximum dry unit weight was measured to be 17 kN/m3. Express your answer as a percentage (but do not write the percentage sign in the answer box).

A cylindrical bar of metal having a diameter of 19.2 mm and a length of 207 mm is deformed elastically in tension with a force of 52900 N. Given that the elastic modulus and Poisson's ratio of the metal are 61.4 GPa and 0.34, respectively, determine the following: a. The amount by which this specimen will elongate in the direction of the applied stress. b. The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.

Reduce the following lambda-calculus term to the normalform. Show all intermediate steps, with one beta reduction at a time. In the reduction, assume that you are supplied with extra rules thatallow you to reduce the multiplication of two natural numbers into thecorresponding result.(λf.λx.f(f x))(λy.y≠3) 2

A hydraulic jump is induced in an 80 ft wide channel. The water depths on either side of the jump are 1 ft and 10 ft. Please calculate: a) The velocity of the faster moving flow. b) The flow rate (discharge). c) The Froude number of the sub-critical flow. d) The flow energy dissipated in the hydraulic jump (expressed as percentage of the energy prior to the jump). e) The critical depth.

A site is compacted in the field, and the dry unit weight of the compacted soil (in the field) is determined to be 18 kN/m3. Determine the relative compaction if the maximum dry unit weight was measured to be 17 kN/m3. Express your answer as a percentage (but do not write the percentage sign in the answer box).

**Explanation:**

One of the common application of debouncing g circuit is in microprocessors or microcontrollers or FPGA's where fast processing is required. In such cases, it is extremely important that during the limited processing cycle, the signals remains valid without debouncinng. Because debouncing can complete impact the output of the controller.

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

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

**B**

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**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)**

**Answer:**

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

**Explanation:**

Knowing

T1 = 25°C = 298 K

T2 = 50°C = 323 K

volume of the tire = V = 0.025

P = 210 kPa (gage)

Pabs = 210 + 101 = 311 KPa

Before the temperature rise

P1 V1 = m1 R1 T1

m1 =

After the temperature rise

P2 =

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

P1 = P2 and V1 = V2

then

m2 =

m2 =

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

P2 = 337.43 KPa

mbleed = 0.007 Kg

Answer:

A) Ductility = 11% EL

B) Radius after deformation = 4.27 mm

Explanation:

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.

Thus;

%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

So;

(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

Answer:

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

Explanation:

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.

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

Also;

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)

Where;

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 =

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.