All categories

3 years ago

Doubling the diameter of a solid, cylindrical wire doubles its strength in tension.

Engineering

1 answer:

julsineya [31]3 years ago

4 0

Answer:

True ❤️

-Solid by solid can make Cylindrical wire doubles Strengths in tension

You might be interested in

Given the system of equations−3x2+7x3=2,x1+2x2–x3=3,5x1−2x2=2(a). Compute the determinant. (b). Use Cramer’s rule to solve for t

34kurt

Answer:

(x1,x2,x3)=(0.985507,1.463768,0.913043)

Explanation:

<em>note: </em>

<em>solution is provided in word document please find the attached document.</em>

Download docx

5 0

3 years ago

What is the definition of a struck by injury

dmitriy555 [2]

Produced by forcible contact or impact between the injured person and an object or a piece of equipment

Explanation:

it mean you get injured after you've already been injured

3 0

3 years ago

Tech A says that multiple gasket or seal leaks can be caused by a defective exhaust gas recirculation (EGR) system.

lesya [120]

I think the answer to this question is b but I’m not sure

7 0

3 years ago

Consider 4.8 pounds per minute of water vapor at 100 lbf/in2, 500 oF, and a velocity of 100 ft/s entering a nozzle operating at

andriy [413]

Answer:

A) $v_2 = 2016.80 ft/s$

B) $\Delta s = 0.006 Btu/lbm R$

Explanation:

Given data:

P-1 = 100 lbf/in^2

$T_1 = 500$ degree f

$V_1 = 100 ft/s$

$P_2 = 40 lbf/inc^2$

effeciency = 80%

from steady flow enerfy equation

$h_1 +\frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}$

where h1 and h2 are inlet and exit enthalpy

for P1 = 100 lbf/in^2 and T1 = 500 degree F

$H_1 = 1278.8 Btu/lbm$

$s_1 = 1.708 Btu/lbm -R$

for P1 = 40 lbf/in^2

$H_1 = 1193.5 Btu/lbm$

$s_1 = 1.708 Btu/lbm -R$

exit enthalapy h_2

$\eta = \frac{h_1 - h'_2}{ h_1 - h_2}$

$0.80 = \frac{1278.8 - h'_2}{1278.8 -1193.5} = 1197.77 Btu/lbm$

from above equation

$1278.8 \times 25037 + \frac{100^2}{2} = 1197.77 \times 25037 + \frac{v_2^2}{2}$ [1 Btu/lbm = 25037 ft^2/s^2]

$v_2 = 2016.80 ft/s$

b) amount of entropy

$\Delta s = s_2 - s_1$

$s_1 = 1.708 Btu/lbm -R$

at $h_2 = 1197.77 Btu/lbm [\tex] and [tex]P_2 = 40 lbf/in^2$

$s_2 is 1.714 Btu/lbm -R$

$\Delta s = 1.714 - 1.708 = 0.006 Btu/lbm R$

6 0

3 years ago

A(n) ____________________ measures the resistance to current flow in a circuit.

yulyashka [42]

Answer: OHMMETER & MEGOHMMETER:

Explanation: The ohmmeter measures circuit resistance; the megohmmeter measures the high resistance of insulation. A meter used to measure electric current. It is connected as part of a circuit.

6 0

3 years ago