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2 years ago

8

A bucket of cement masse 40kg is tied to the end of the rope connected to a hoist.calculate the tension in the rope when the buc

ket is suspended but stationery

1 answer:

Alik [6]2 years ago

4 0

The tension in the rope when this bucket is suspended but stationery (static) is equal to 400 Newton.

<u>Given the following data:</u>

Mass = 40 kg.

<h3>How to calculate the tension?</h3>

Mathematically, the tension in the rope when this bucket is suspended but stationery (static) can be calculated by using this formula:

T = m(a + g)

<u>Note:</u> Acceleration due to gravity (g) is equal to 10 m/s² while the acceleration is 0 m/s² because the bucket is stationery (static)

Substituting the parameters into the formula, we have;

T = 40(0 + 10)

T = 400 Newton.

Read more on tension here: brainly.com/question/4080400

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In a tensile test on a steel specimen, true strain = 0.12 at a stress of 250 MPa. When true stress = 350 MPa, true strain = 0.26

scZoUnD [109]

Answer:

The strength coefficient is $625$ and the strain-hardening exponent is $0.435$

Explanation:

Given the true strain is 0.12 at 250 MPa stress.

Also, at 350 MPa the strain is 0.26.

We need to find $(K)$ and the $(n)$ .

$\sigma =K\epsilon^n$

We will plug the values in the formula.

$250=K\times (0.12)^n\\350=K\times (0.26)^n$

We will solve these equation.

$K=\frac{250}{(0.12)^n}$ plug this value in $350=K\times (0.26)^n$

$350=\frac{250}{(0.12)^n}\times (0.26)^n\\ \\\frac{350}{250}=\frac{(0.26)^n}{(0.12)^n}\\ \\1.4=(2.17)^n$

Taking a natural log both sides we get.

$ln(1.4)=ln(2.17)^n\\ln(1.4)=n\times ln(2.17)\\n=\frac{ln(1.4)}{ln(2.17)}\\ n=0.435$

Now, we will find value of $K$

$K=\frac{250}{(0.12)^n}$

$K=\frac{250}{(0.12)^{0.435}}\\ \\K=\frac{250}{0.40}\\\\K=625$

So, the strength coefficient is $625$ and the strain-hardening exponent is $0.435$ .

5 0

3 years ago

Will Give Brainliest

Tom [10]

Answer:

A hole within a hole

Explanation:

8 0

3 years ago

Why is it important to know the accuracy and precision of a measuring device? Do you think that the dial caliper manufacturer’s

Leto [7]

Answer:

Accuracy and precision allow us to know how much we can rely on a measuring device readings. ±.001 as a "accuracy" claim is vague because there is no unit next to the figure and the claim fits better to the definition of precision.

Explanation:

Accuracy and Precision: the golden couple.

Accuracy and precision are key elements to define if a measuring device is reliable or not for a specific task. Accuracy determines how close are the readings from the ideal/calculated values. On the other hand, precision refers to repeatability, that is to say how constant the readings of a device are when measuring the same element at different times. One of those two key concepts may not fulfill the criteria for measuring tool to be used on certain engineering projects where lack of accuracy (disntant values from real ones) or precision (not constant readings) may lead to malfunctons and severe delays on the project development.

±.001 what unit?

The manufacturer says that is an accuracy indicator, nevertheless there is now unit stated so this is not useful to see how accurate the device is. Additionally, That notation is more used to refer to device tolerances, that is to say the range of possible values the instrument may show when reading and element. It means it tells us more about the device precision during measurments than actual accuracy. I would recommend the following to the dial calipers manufacturers to better explain its measurement specifications:

Use ±.001 as a reference for precision. It is important to add the respective unit for that figure.
Condcut test to define the actual accuracy value an present it using one of the common used units for that: Error percentage or ppm.

3 0

3 years ago

Please help meeeee. I don’t have time

Sholpan [36]

Answer: precision

Explanation: because accuracy is right on there and precision is getting closer and closer

3 0

3 years ago

Water flows in a constant diameter pipe with the following conditions measured:

Burka [1]

Answer:

a) $h_L=-3.331ft$

b) The flow would be going from section (b) to section (a)

Explanation:

1) Notation

$p_a =31.1psi=4478.4\frac{lb}{ft^2}$

$p_b =27.3psi=3931.2\frac{lb}{ft^2}$

For above conversions we use the conversion factor $1psi=144\frac{lb}{ft^2}$

$z_a =56.7ft$

$z_a =68.8ft$

$h_L =?$ head loss from section

2) Formulas and definitions

For this case we can apply the Bernoulli equation between the sections given (a) and (b). Is important to remember that this equation allows en energy balance since represent the sum of all the energies in a fluid, and this sum need to be constant at any point selected.

The formula is given by:

$\frac{p_a}{\gamma}+\frac{V_a^2}{2g}+z_a =\frac{p_b}{\gamma}+\frac{V_b^2}{2g}+z_b +h_L$

Since we have a constant section on the piple we have the same area and flow, then the velocities at point (a) and (b) would be the same, and we have just this expression:

$\frac{p_a}{\gamma}+z_a =\frac{p_b}{\gamma}+z_b +h_L$

3)Part a

And on this case we have all the values in order to replace and solve for $h_L$

$\frac{4478.4\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+56.7ft=\frac{3931.2\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+68.8ft +h_L$

$h_L=(71.769+56.7-63-68.8)ft=-3.331ft$

4)Part b

Analyzing the value obtained for $\h_L$ is a negative value, so on this case this means that the flow would be going from section (b) to section (a).

5 0

3 years ago