A __________ defines the area of land that will supply water to a stream.

What is the stress concentration factor of a shaft in torsion, where D=1.25 in. and d=1 in. and the fillet radius is, r=0.2 in.a

klio [65]

Answer:

Concentration factor will be 1.2

So option (C) will be correct answer

Explanation:

We have given outer diameter D = 1.25 in

And inner diameter d = 1 in and fillet ratio r = 0.2 in

So $\frac{r}{d}$ ratio will be $=\frac{0.2}{1}=0.2$

And $\frac{D}{d}$ ratio will be $=\frac{1.25}{1}=1.25$

Now from the graph in shaft vs torsion the value of concentration factor will be 1.2

So concentration factor will be 1.2

So option (C) will be correct answer.

6 0

2 years ago

3. (5%) you would like to physically separate different materials in a scrap recycling plant. describe at least one method that

Likurg_2 [28]

One of the methods that are used to separate polymers, aluminium alloys, and steels from one another is the Gravitation Separation method.

One straightforward technique is to run the mixture through a magnet, which will keep the steel particles on the magnet and separate them from the polymer.

What is the Gravitation Separation method?

When it is practicable to separate two components using gravity, i.e., when the combination's constituent parts have different specific weights, gravity separation is a technique used in industry. The components can be in suspension or in a dry granular mixture.

Polymers, Steel and Aluminium alloys can be readily split apart. The technique depends on how the two components are combined. The approach used is gravitational density. Due to the significant difference in relative specific mass values between steel and polymers (which range from 1.0 to 1.5), it is possible to separate them using flotation in a liquid that is safe and has the right density.

Therefore, the Gravitation Separation method is used to separate polymers, aluminium alloys and steels.

To learn more about the Polymer from the given link

brainly.com/question/2494725

#SPJ4

8 0

11 months ago

A cast-iron tube is used to support a compressive load. Knowing that E 5 10 3 106 psi and that the maximum allowable change in l

Papessa [141]

Answer:

(a) 2.5 ksi

(b) 0.1075 in

Explanation:

(a)

$E=\frac {\sigma}{\epsilon}$

Making $\sigma$ the subject then

$\sigma=E\epsilon$

where $\sigma$ is the stress and $\epsilon$ is the strain

Since strain is given as 0.025% of the length then strain is $\frac {0.025}{100}=0.00025$

Now substituting E for $10\times 10^{6} psi$ then

$\sigma=(10\times 10^{6} psi)\times 0.00025=2500 si= 2.5 ksi$

(b)

Stress, $\sigma= \frac {F}{A}$ making A the subject then

$A=\frac {F}{\sigma}$

$A=\frac {\pi(d_o^{2}-d_i^{2})}{4}$

where d is the diameter and subscripts o and i denote outer and inner respectively.

We know that $2t=d_o - d_i$ where t is thickness

Now substituting

$\frac {\pi(d_o^{2}-d_i^{2})}{4}=\frac {1600}{2500}$

$\pi(d_o^{2}-d_i^{2})=\frac {1600}{2500}\times 4$

$(d_o^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4$

But the outer diameter is given as 2 in hence

$(2^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4$

$2^{2}-(\frac {1600}{2500\times \pi}\times 4)=d_i^{2}$

$d_i=\sqrt {2^{2}-(\frac {1600}{2500\times \pi}\times 4)}=1.784692324 in\approx 1.785 in$

As already mentioned, $2t=d_o - d_i hence t=0.5(d_o - d_i)$

$t=0.5(2-1.785)=0.1075 in$

3 0

3 years ago

Please help me with this question

kap26 [50]

Answer:

The current through each lamp is 0.273 Amperes

Power dissipated in each lamp is 0.082W

Explanation:

Battery v = 1.5 V

Each lamp has resistance, r = 1.1 Ohms

The 5 lamps in series will therefore have total resistance, R = 5 * 1.1 = 5.5 Ohms

The current through each lamp, I = v/R = 1.5 / 5.5 = 0.273 Amperes

Power dissipated in each lamp = I² * r = 0.273² * 1.1 = 0.082W

3 0

2 years ago

A 1000 KVA three phase transformer has a secondary voltage of 208/120. What is the secondary full load amperage?

IceJOKER [234]

Answer:

The three phase full load secondary amperage is 2775.7 A

Explanation:

Following data is given,

S = Apparent Power = 1000 kVA

No. of phases = 3

Secondary Voltage: 208 V/120 V (Here 208 V is three phase voltage and 120 V is single phase voltage)

Since,

$V_{1ph} =\frac{ V_{3ph}}{\sqrt{3} }\\V_{1ph) = \frac{208}{\sqrt{3} }\\$

$V_{1ph} = 120 V$

The formula for apparent power in three phase system is given as:

$S = \sqrt{3} VI$

Where:

S = Apparent Power

V = Line Voltage

I = Line Current

In order to calculate the Current on Secondary Side, substituting values in above formula,

$1000 kVA = \sqrt{3} * (208) * (I)\\1000 * 1000 = \sqrt{3} * (208) * (I)\\I = \frac{1000 * 1000}{\sqrt{3} * (208) }\\ I = 2775.7 A$

4 0

2 years ago