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EleoNora [17]
3 years ago
14

A car travells at 67.5 km\h in 120 km.how long will it take to reach the destination

Engineering
1 answer:
Kruka [31]3 years ago
5 0

mark me the brainiest here

average speed (in km/h) of a car stuck in traffic that drives 12 kilometers in 2 hours.

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11) (10 points) A large valve is to be used to control water supply in large conduits. Model tests are to be done to determine h
IrinaVladis [17]

Answer:

7.94 ft^3/ s.

Explanation:

So, we are given that the '''model will be 1/6 scale (the modeled valve will be 1/6 the size of the prototype valve)'' and the prototype flow rate is to be 700 ft3 /s. Then, we are asked to look for or calculate or determine the value for the model flow rate.

Note that we are to use Reynolds scaling for the velocity as par the instruction from the question above.

Therefore; kp/ks = 1/6.

Hs= 700 ft3 /s and the formula for the Reynolds scaling => Hp/Hs = (kp/ks)^2.5.

Reynolds scaling==> Hp/ 700 = (1/6)^2.5.

= 7.94 ft^3/ s

7 0
3 years ago
For which of 'water' flow velocities at 200C can we assume that the flow is incompressible ? a.1000 km per hour b. 500 km per ho
ad-work [718]

Answer:d

Explanation:

Given

Temperature=200^{\circ}\approc 473 K

Also \gamma for air=1.4

R=287 J/kg

Flow will be In-compressible when Mach no.<0.32

Mach no.=\frac{V}{\sqrt{\gamma RT}}

(a)1000 km/h\approx 277.78 m/s

Mach no.=\frac{277.78}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.63

(b)500 km/h\approx 138.89 m/s

Mach no.=\frac{138.89}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.31

(c)2000 km/h\approx 555.55 m/s

Mach no.=\frac{555.55}{\sqrt{1.4\times 287\times 473}}

Mach no.=1.27

(d)200 km/h\approx 55.55 m/s

Mach no.=\frac{55.55}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.127

From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.

5 0
3 years ago
An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e
Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

3 0
3 years ago
What type of engineer makes sure equipment is safe and operational
zvonat [6]

Answer:

mechanical engineer is the best answer

8 0
3 years ago
A 30 mm thick AISI 1020 steel plate is sandwiched between two 10 mm thick 2024-T3 aluminum plates and compressed with a bolt and
denis-greek [22]

Answer:

275 MPa

Explanation:

Regardless of what it is holding, the stiffness of a bolt depends on its own material properties and geometry.

The stiffness is:

k = E * \frac{A}{l}

I assume this one is made of steel, because regular bolts are steel.

The Young's modulus for steel is E = 210 GPa

The longitude is given. (But note that in a real application you have to consider the length up to the nut.)

The section is (using the nominal diameter of 10 mm)

A = \frac{\pi * d^2}{4} = \frac{\pi * 0.01^2}{4} = 7.85e-5 m^2

Then:

k  = 2.1e11 * \frac{7.85e-5}{0.06} = 275e6 Pa = 275 MPa

5 0
3 years ago
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