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larisa86 [58]
3 years ago
11

. A normal-weight concrete has an average compressive strength of 20 MPa. What is the estimated flexure strength

Engineering
1 answer:
bulgar [2K]3 years ago
4 0

Answer:

2.77mpa

Explanation:

compressive strength = 20 MPa. We are to find the estimated flexure strength

We calculate the estimated flexural strength R as

R = 0.62√fc

Where fc is the compressive strength and it is in Mpa

When we substitute 20 for gc

Flexure strength is

0.62x√20

= 0.62x4.472

= 2.77Mpa

The estimated flexure strength is therefore 2.77Mpa

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Answer:

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Explanation:

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3 years ago
In a parallel one-dimensional flow in the positive x direction, the velocity varies linearly from zero at y = 0 to 32 m/s at y =
monitta

Answer:

Ψ = 10(y^2) + c

<em><u>y = 1.067m</u></em>

Explanation:

since the flow is one dimensional in positive X direction, the only velocity component is in X, which is denoted by u

while u is a function of y

we find the u in terms of y; u varies linearly wih y

we use similiraty to find the relation

32/1.6 =<em>u/y</em>

<em><u>u = 20y</u></em>

<em><u>Ψ = ∫20ydy</u></em>

<em><u>Ψ = 10(y^2) + c</u></em>

<em><u>(b)</u></em>

<em><u>the flow is half below y = 1.6*(2/3)=1.067 m</u></em>

<em><u>this is because at two third of the height of a triangle lies the centroid of triangle. since the velocity profile forms a right angled triangle , its height is 1.6 m . the flow is halved at y = 1.067m</u></em>

3 0
3 years ago
Other examples of joption​
Evgen [1.6K]

Answer:

Profit from stock price gains with limited risk and lower cost than buying the stock outright.

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A particle moving in the x-y plane has a position vector given by r = 1.89t2i + 1.17t3j, where r is in inches and t is in second
ehidna [41]

Answer:

r=89.970 m

Explanation:

We need to calculate radius of curvature, using first and second derivatives of the functions x(t) and y(t).

From the given equation, we can get, that:

x(t)=1.89t^2 and y(t)=1.17t^3

Then, the first derivative:

x'(t)=2*1.89*t=3.78t and y'(t)=3*1.17*t^2=3.51t^2

The second derivatives are:

x''(t)=3.78 and y''(t)=3.51*2*t=7.02t

Equation for the radius of curvature, can be found as:

r=(x'^2+y'^2)^(3/2)/(x''y'-x'y'')

Note, that the denominator should be taken by its absolute value.

For the given time, we should calculate numerical values for the derivatives. For t=2.1 s:

x'(2.1)=7.938 m/s; x''(2.1)=3.78 m/s^2; y'(2.1)=15.4791 m/s and y''(2.1)=14.742 m/s^2

Using equation of the curvature radius and the values, we can get:

r=89.970 m

6 0
3 years ago
Read 2 more answers
An automobile engine provides 632 Joules of work to push the pistons and generates 2203 Joules of heat that must be carried away
Ronch [10]

Answer:

The change in internal energy of the engine is -1,571 Joules.

Explanation:

Change in internal energy (∆U) = energy output (U2) - energy generated (U1)

U1 is the energy generated = 2203 Joules

U2 is the energy output = 632 Joules

Change in internal energy (∆U) = 632 - 2203 = -1,571 Joules

3 0
3 years ago
Read 2 more answers
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