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ivann1987 [24]
3 years ago
10

Rockfalls can cause major damage to roads and infrastructure. To design mitigation bridges and barriers, engineers use the coeff

icient of restitution to model the behavior of the rocks. The rock A falls a distance of 32 m before striking an incline with slope α = 40°. Knowing that the coefficient of restitution between A and the incline is 0.2, determine the velocity of the rock after the impact.

Engineering
1 answer:
Rus_ich [418]3 years ago
5 0

Answer:

Velocity of rock fall collision, v = 13.21 m/sec

Explanation:

Detailed explanation given in the attached document.

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A company wants to develop audio speakers using an inexpensive type of plastic that has a very high quality of sound output. Whi
topjm [15]

Answer:A

Explanation:

Those who want to save money and will use the product for only a few years

5 0
3 years ago
Calculate the convective heat-transfer coefficient for water flowing in a round pipe with an inner diameter of 3.0 cm. The water
olasank [31]

Answer:

h = 10,349.06 W/m^2 K

Explanation:

Given data:

Inner diameter = 3.0 cm

flow rate  = 2 L/s

water temperature 30 degree celcius

Q = A\times V

2\times 10^{-3} m^3 = \frac{\pi}{4} \times (3\times 10^{-2})^2 \times velocity

V = \frac{20\times 4}{9\times \pi} = 2.83 m/s

Re = \frac{\rho\times V\times D}{\mu}

at 30 degree celcius = \mu = 0.798\times 10^{-3}Pa-s , K  = 0.6154

Re = \frac{10^3\times 2.83\times 3\times 10^{-2}}{0.798\times 10^{-3}}

Re = 106390

So ,this is turbulent flow

Nu = \frac{hL}{k} = 0.0029\times Re^{0.8}\times Pr^{0.3}

Pr= \frac{\mu Cp}{K} = \frac{0.798\times 10^{-3} \times 4180}{0.615} = 5.419

\frac{h\times 0.03}{0.615}  = 0.0029\times (1.061\times 10^5)^{0.8}\times 5.419^{0.3}

SOLVING FOR H

WE GET

h = 10,349.06 W/m^2 K

6 0
3 years ago
You want to improve your grades so that you make all A's next 6 weeks. List examples of quantitative and qualitative data you sh
Naddika [18.5K]

Explanation:

z3d33sxurljt 36f

3fभथठभदाफमदखज्ञफादफज्ञादफज्ञिलफ इऋबिअऋब

6 0
3 years ago
An asbestos pad is square in cross section, measuring 5 cm on a side at its small end increasing linearly to 10 cm on a side at
polet [3.4K]

Answer:

q = 1.73 W

Explanation:

given data

small end  = 5 cm

large end = 10 cm

high = 15 cm

small end is held = 600 K

large end at = 300 K

thermal conductivity of asbestos  = 0.173 W/mK

solution

first we will get here side of cross section that is express as

S = S1 + \frac{S2-S1}{L} x     ...............1

here x is distance from small end and S1 is side of square at small end

and S2 is side of square of large end and L is length

put here value and we get

S = 5 + \frac{10-5}{15} x

S = \frac{0.15 + x}{3}    m

and  

now we get here Area of section at distance x is

area A = S²    ...............2

area A = (\frac{0.15 + x}{3})^2    m²

and

now we take here small length dx and temperature difference is dt

so as per fourier law

heat conduction is express as

heat conduction q = \frac{-k\times A\  dt}{dx}      ...............3

put here value and we get

heat conduction q = -k\times (\frac{0.15 + x}{3})^2 \   \frac{dt}{dx}  

it will be express as

q \times \frac{dx}{(\frac{0.15 + x}{3})^2} = -k (dt)  

now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K

q \int\limits^{0.15}_0 {\frac{dx}{(\frac{0.15 + x}{3})^2 } = -0.173 \int\limits^{300}_{600} {dt}          

solve it and we get

q (30)  = (0.173) × (600 - 300)

q = 1.73 W

5 0
3 years ago
How do i get a used eraser out of a mechanical pencil
timurjin [86]

Answer:

Use a screw or corkscrew.  Screw it into the eraser, then pull.

4 0
3 years ago
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