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Finger [1]
2 years ago
10

A student used a 500-ml graduated cylinder to measure the volume of water in a 1-cup measure. three trials of the measurement ga

ve volumes of 240 ml, 242 ml, and 235 ml. What is the average of the three measurements?
Engineering
1 answer:
Sedaia [141]2 years ago
5 0

Answer:

  239 mL

Explanation:

The average of any set of data is the sum of the values, divided by their number.

  v = (240 +242 +235)/3 = 717/3 = 239 . . . mL

The average of the three measurements is 239 mL.

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Motion is defined as a change in an object's position when compared to other objects around it. Mary Ann was watching a slug cra
lara [203]

Answer:

Explanation:

Do it yourself bum

4 0
3 years ago
Given a program with execution times broken down shown below. Assume that techniques can only be applied to accelerate the integ
Alex17521 [72]

<u>Solution and Explanation:</u>

Floating point and Integer consumes 1000 seconds of 1600 seconds.

Since we shall be talking of theoretical speed-up, please do not try to relate this with any practical scenario :P

To achieve maximum speed up, we need to reduce the time consumed by FP and Integer as large as possible. So we consider a system that does not at all consume time to perform FP and Integer operations. With such specifications, we can say that our new system will consume 600 seconds.

 speed up = 1600 divide by 600 = 2.66

<u>part b: </u>

We need a System with enhancement that will result in speedup of 2.

So the time required for the new system would be 800 sec

 Required time = 1600 divide by = 800 seconds

It is now given that Floating Point can now be accelerated by 5 times, so our enhanced system will consume 40 seconds to perform Floating point operations

We know that Load/Store and branch operations cannot be enhanced and hence they will consume 600 seconds.

Therefore to attend speedup of 2, Integer operations must be completed in 160 seconds (160 = 800 – (600 + 40))

So speed up required for Integer Operations is 800/160 = 5

So Integer Operation should need 5 times less time to achieve speed up of 2

7 0
3 years ago
Consider a fully developed laminar flow in a circular pipe. The velocity at R/2 (midway between the wall surface and the centerl
ryzh [129]

Answer:

The velocity at R/2 (midway between the wall surface and the centerline) is given by (3/4)(Vmax) provided that Vmax is the maximum velocity in the tube.

Explanation:

Starting from the shell momentum balance equation, it can be proved that the velocity profile for fully developedblaminar low in a circular pipe of internal radius R and a radial axis starting from the centre of the pipe at r=0 to r=R is given as

v = (ΔPR²/4μL) [1 - (r²/R²)]

where v = fluid velocity at any point in the radial direction

ΔP = Pressure drop across the pipe

μ = fluid viscosity

L = pipe length

But the maximum velocity of the fluid occurs at the middle of the pipe when r=0

Hence, maximum veloxity is

v(max) = (ΔPR²/4μL)

So, velocity at any point in the radial direction is

v = v(max) [1 - (r²/R²)]

At the point r = (R/2)

r² = (R²/4)

(r²/R²) = r² ÷ R² = (R²/4) ÷ (R²) = (1/4)

So,

1 - (r²/R²) = 1 - (1/4) = (3/4)

Hence, v at r = (R/2) is given as

v = v(max) × (3/4)

Hope this Helps!!!

4 0
2 years ago
Quelles sont les types de carburant utilisés en aviation
zhenek [66]

hope it's help you ok have a good day

6 0
2 years ago
A civil engineer is asked to design a curved section of roadway that meets the following conditions: With ice on the road, when
lianna [129]

Answer:

1. 3.4^{o}

2. 163.3 m

Explanation:

Static friction between road and rubber, μs =0.06

The maximum speed of the car, v = 50 km/h

                                              = (50)(1000/3600) m/s

                                               = 13.89 m/s

The acceleration due to gravity, g = 9.81 m/s^{2}

The frictional force, f = μsN     ...... (1)

The component mg cosθ which balance the normal reaction N

The component mg sinθ acts in an opposite direction to the frictional force f.

        ΣF = mg sinθ-f = 0      ...... (2)

Substitute the equation (1) in equation (2), we get

 ΣF = mgsinθ-μsN = 0

 mgsinθ-μsmgcosθ =0

 μs = sinθ/cosθ

   tanθ = μs

    θ = tan-1( μs) = tan-1(0.06) = 3.4^{o}

(b)The vertical component of the force is

N cosθ = fsinθ+mg

 N cosθ = μsNsinθ+mg

N[cosθ- μs sinθ] = mg     ...... (3)

The horizontal component of the force along the motion of the car is

Nsinθ+fcosθ = ma  (Centripetal acceleration, a = \frac {v^{2}}{r}

  Nsinθ+fcosθ = m(\frac {v^{2}}{r})

   Nsinθ+μsNcosθ = m(\frac {v^{2}}{r})

N[sinθ+μs cosθ] = m(\frac {v^{2}}{r})     ...... (4)    

Dividing the equation (4) with equation (3),

 [sinθ+μscosθ]/[cosθ- μs sinθ] = \frac {v^{2}}{rg}

 cosθ[sinθ/cosθ+μs]/cosθ[1- μs sinθ/cosθ] =\frac {v^{2}}{rg}

[tanθ+μs]/[1-μs tanθ] = \frac {v^{2}}{rg}      

 From part (1), tanθ = μs

 Then the above equation becomes

 \frac {(\mu_s+\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}

\frac {(2\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}

Therefore, the minimum radius of the curvature of the curve is

               r = \frac {v^{2}}{{2 \mu_s/[1-\mu_s^{2}]}g} 

                   = \frac {v^{2}[1-\mu_s^{2}]}{2\mu_s g}

                   = \frac {(13.89 m/s)^{2}[1-(0.06)^{2}]}{(2)(0.06)(9.81)}

                 = 163.3 m

5 0
3 years ago
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