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marysya [2.9K]
3 years ago
7

Write a program. Submit a file named height.c Amusement parks have minimum heights for their rides. Create a program that adds u

sers to the line of a ride in an amusement park if they meet the minimum height requirement (given in inches) . Create three classes: o An amusement park class o A ride class o Arider class (people riding the rides) It should work with the following main (DO NOT MODIFY-NO CREDIT IF YOU MODIFY): int mainO Rider r1(Yaris", 45); //name, height in inches Rider r2(49)://height in inches Amusement park a1(3); //3 is the number of rides in the amusement park a1.get ride(1).add_line(r1); //add a rider to the line of a ride Amusement park a2(2); //2 is the number of rides in the amusement park a2.get ride(1).add line(r2); //add a rider to the line of a ride return 0; Possible Sample Run 1 (vour program does not have to match the following exactlv, but ~~~Amusement Park into~~~ Ride 1- Enter minimum ride height and ride name: 40 ride1 Ride 2- Enter minimum ride height and ride name 50 ride2 Ride 3- Enter minimum ride height and ride name 60 ride3 Adding rider to line Amusement Park Info Ride 1- Enter minimum ride height and ride name: 75 ride1 Ride 2- Enter minimum ride height and ride name: 76 ride2 Sorry can't add rider-too short.
Engineering
1 answer:
Natasha_Volkova [10]3 years ago
8 0

Answer:

you ok

Explanation:

try it

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A stainless steel ball (=8055 kg/m3, Cp= 480 J/kgK) of diameter D =15 cm is removed from theoven at a uniform temperature of 3
aleksandrvk [35]

Answer:

i) 25.04 W/m^2 .k

ii) 23.82 minutes = 1429.2 secs

Explanation:

Given data:

Diameter of steel ball = 15 cm

uniform temperature = 350°C

p = 8055 kg/m^3

Cp = 480 J/kg.k

surface temp of ball drops to  250°C

average surface temperature = ( 350 + 250 ) / 2 = 300°C

<u>i) Determine the average convection heat transfer coefficient during the cooling process</u>

<em>Note : Obtain the properties of air at 1 atm at average film temp of 30°C from the table  " properties of air "  contained in your textbook</em>

average convection heat transfer coefficient = 25.04 W/m^2 .k

<u>ii) Determine how long this process has taken </u>

Time taken by the process = 23.82 minutes = 1429.2 seconds

Δt = Qtotal / Qavg = 683232 / 477.92 = 1429.59 secs

attached below is the detailed solution of the given question

3 0
3 years ago
a ball is subject to two forces F1 and F2. The magnitudes of the two forces are 45.0 N and 70.0 N respectively. In the figure be
tino4ka555 [31]

Answer:

  • F1.x ≈ -28.93
  • F1.y ≈ 34.47
  • F2.x = 70
  • F2.y = 0
  • (F1+F2).x ≈ 41.07
  • (F1+F2).y ≈ 34.47
  • |F1+F2| ≈ 53.62
  • ∠(F1+F2) ≈ 40.0°

Explanation:

A suitable calculator can show you the vector components and their resultant in polar or rectangular format. (See attached.) 2D vectors are conveniently treated as complex numbers, which is why the y-component values are shown as imaginary.

(The 50° angle measured from the -x axis is equivalent to 130° measured from the +x axis, which is the reference we're using here.)

If you'd like to compute the vector components by hand, they are ...

  (x, y) = magnitude×(cos(angle), sin(angle))

This notation is sometimes abbreviated <em>magnitude cis angle</em>, a reference to the complex number form x+yi.

8 0
2 years ago
Technician a says that diesel engines can produce more power because air in fuel or not mix during the intake stroke. Technician
mariarad [96]

Answer:

Technician be says that diesel engines produce more power because they use excess air to burn feel who is correct

Explanation:

He is correct as many engines are run by diesel. It produces more power as that is how cars produce more power.

3 0
3 years ago
Show the bias polarities and depletion regions of an npn BJT in the normal active, saturation, and cutoff modes of operation. Dr
Troyanec [42]

Complete Question:

Show the bias polarities and depletion regions of an npn BJT in the normal active, saturation, and cutoff modes of operation. Draw the three sketches one below the other to (qualitatively) reflect the depletion widths for these biases, and the relative emitter, base, and collector doping.

Consider a BJT with a base transport factor of 1.0 and an emitter injection efficiency of 0.5.

Calculate roughly by what factor would doubling the base width of a BJT would increase, decrease, or leave unchanged the emitter injection efficiency and base transport factor? Repeat for the case of emitter doping increased 5 × =. Explain with key equations, and assume other BJT parameters remain unchanged!

Answer & Explanation:

[Find the attachments]

Step 1 :

Emitter and base, collector, and base are forward biased then BJT is in saturation region. Emitter and base is forward biased and base and collector in reverse biased then BJT is in active region.

Emitter and base, collector and base are reverse biased then BJT in cut off region.

Three sketches one below the other is shown in Figure 1.

[find the figure in attachment]

Step 2:

Value of base widths of saturation, active and cut off operated BJT are value of Base width of saturated region operated BJT is less than base width in active region operated BJT. Value of base width of active region operated BJT is less than base width in cut off region operated BJT.

Saturation region operated base width of BJT is < Active region operated base width of BJT is < Cut off region operated base width of BJT.

[For  Steps 3 4 5 6 and 7 find attachments]

8 0
4 years ago
A cylindrical rod 100 mm long and having a diameter of 10.0 mm is to be deformed using a tensile load of 27,500 N. It must not e
mash [69]

Answer:

The steel is a candidate.

Explanation:

Given

P = 27,500 N

d₀ = 10.0 mm = 0.01 m

Δd = 7.5×10 ⁻³ mm (maximum value)

This problem asks that we assess the four alloys relative to the two criteria presented. The first  criterion is that the material not experience plastic deformation when the tensile load of 27,500 N is  applied; this means that the stress corresponding to this load not exceed the yield strength of the material.

Upon computing the stress

σ = P/A₀ ⇒ σ = P/(π*d₀²/4) = 27,500 N/(π*(0.01 m)²/4) = 350*10⁶ N/m²

⇒ σ = 350 MPa

Of the alloys listed, the Ti and steel alloys have yield strengths greater than 350 MPa.

Relative to the second criterion, (i.e., that Δd be less than 7.5 × 10 ⁻³  mm), it is necessary to  calculate the change in diameter Δd for these four alloys.

Then we use the equation   υ = - εx / εz = - (Δd/d₀)/(σ/E)

⇒  υ = - (E*Δd)/(σ*d₀)

Now, solving for ∆d from this expression,

∆d = - υ*σ*d₀/E

For the Aluminum alloy

∆d = - (0.33)*(350 MPa)*(10 mm)/(70*10³MPa) = - 0.0165 mm

0.0165 mm > 7.5×10 ⁻³ mm

Hence, the Aluminum alloy is not a candidate.

For the Brass alloy

∆d = - (0.34)*(350 MPa)*(10 mm)/(101*10³MPa) = - 0.0118 mm

0.0118 mm > 7.5×10 ⁻³ mm

Hence, the Brass alloy is not a candidate.

For the Steel alloy

∆d = - (0.3)*(350 MPa)*(10 mm)/(207*10³MPa) = - 0.005 mm

0.005 mm < 7.5×10 ⁻³ mm

Therefore, the steel is a candidate.

For the Titanium alloy

∆d = - (0.34)*(350 MPa)*(10 mm)/(107*10³MPa) = - 0.0111 mm

0.0111 mm > 7.5×10 ⁻³ mm

Hence, the Titanium alloy is not a candidate.

7 0
3 years ago
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