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Marizza181 [45]
3 years ago
8

A steel loop ABCD of length 5ft and of 3/8" diameter is placed as shown around a 1" diameter aluminum rod AC. Cables BE and DF e

ach 1/2" diameter are used to apply the load Q. Knowing that the ultimate strength of the steel used for the loop and the cables is 70ksi and that the ultimate strength of the aluminum used for the rod is 38ksi. Determine the largest load Q that can be applied if an overall factor of safety of 3 is desired.
Engineering
1 answer:
lorasvet [3.4K]3 years ago
4 0

Answer:

a

Explanation:

ye men thats the answer<em><u>#</u></em><em><u>#</u></em><em><u>#</u></em><em><u>#</u></em><em><u>#</u></em><em><u>#</u></em><em><u>#</u></em><em><u>%</u></em>

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How fast is a 2012 nissan sentra<br>speed and acceleration ​
Alika [10]

Answer:

it has 15 horsepower to 300 horsepower and it weighs 2,906 to 3,131

Explanation:

its torque is 142 to 180

it has a inline 4 engine

there's a SE-R which has a turbo

4 0
3 years ago
Let A→=(150iˆ+270jˆ) mm , B→=(300iˆ−450jˆ) mm , and C→=(−100iˆ−250jˆ) mm . Find scalars r and s, if possible, such that R→=rA→+s
ioda

Answer: r = 0.8081; s = -0.07071

Explanation:

A = (150i + 270j) mm

B = (300i - 450j) mm

C = (-100i - 250j) mm

R = rA + sB + C = 0i + 0j

R = r(150i + 270j) + s(300i - 450j) + (-100i - 250j) = 0i + 0j

R = (150r + 300s - 100)i + (270r - 450s - 250)j = 0i + 0j

Equating the i and j components;

150r + 300s - 100 = 0

270r - 450s - 250 = 0

150r + 300s = 100

270r - 450s = 250

solving simultaneously,

r = 0.8081 and s = -0.07071

QED!

5 0
4 years ago
What must engineers keep in mind so that their solutions will be appropriate?
vekshin1

Answer:

Context

Explanation:

It is of great value for an engineer to keep the context of his/her experiment in mind.

7 0
3 years ago
Reverse masking forms a soft edge on the panel.
valina [46]
True I think I’m not sure?
7 0
2 years ago
In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles o
lesya [120]

Answer:

The molecular weight will be "28.12 g/mol".

Explanation:

The given values are:

Pressure,

P = 10 atm

  = 10\times 101325 \ Pa

  = 1013250 \ Pa

Temperature,

T = 298 K

Mass,

m = 11.5 Kg

Volume,

V = 1000 r

   = 1 \ m^3

R = 8.3145 J/mol K

Now,

By using the ideal gas law, we get

⇒ PV=nRT

o,

⇒ n=\frac{PV}{RT}

By substituting the values, we get

       =\frac{1013250\times 1}{8.3145\times 298}

       =408.94 \ moles

As we know,

⇒ Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}

or,

⇒        MW=\frac{m}{n}

                   =\frac{11.5}{408.94}

                   =0.02812 \ Kg/mol

                   =28.12 \ g/mol

3 0
3 years ago
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