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3 years ago

What is the perimeter of 14-7 and 3-4

Engineering

1 answer:

Goshia [24]3 years ago

5 0

Answer:

If you mean two sides are 7 and two sides are 14 then you'd have 42

and for the second you'd have 14

Explanation:

7 + 7 = 14, 14 + 14 = 28, 14 + 28 = 42

3 + 3 = 6, 4 + 4 = 8, 8 + 6 = 14

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A plate clutch has a single friction surface 9-in OD by 7-in ID. The coefficient of friction is 0.2 and the maximum pressure is

Talja [164]

Answer:

the torque capacity is 30316.369 lb-in

Explanation:

Given data

OD = 9 in

ID = 7 in

coefficient of friction = 0.2

maximum pressure = 1.5 in-kip = 1500 lb

To find out

the torque capacity using the uniform-pressure assumption.

Solution

We know the the torque formula for uniform pressure theory is

torque = 2/3 × $\pi$ × coefficient of friction × maximum pressure ( R³ - r³ ) .....................................1

here R = OD/2 = 4.5 in and r = ID/2 = 3.5 in

now put all these value R, r, coefficient of friction and maximum pressure in equation 1 and we will get here torque

torque = 2/3 × $\pi$ × 0.2 × 1500 ( 4.5³ - 3.5³ )

so the torque = 30316.369 lb-in

3 0

3 years ago

three string are attached to a small metal ring, two of the strings make and angle of 35° with the vertical and each is pulled w

Julli [10]

No clue sorry man I would help but I need help too

6 0

2 years ago

A 75 ! coaxial transmission line has a length of 2.0 cm and is terminated with a load impedance of 37.5 j75 !. If the relative p

Cerrena [4.2K]

Answer:

4.26

Explanation:

The wavelength λ is given by:

$\lambda=v/f=c/nf\\c=speed\ of\ light=3*10^8m/s,f=frequency=3*10^9Hz,n=permittivity=2.56\\\\\lambda=3*10^8/(2.56*3*10^9)=0.0625\ m\\$

Phase constant (β) = 2π/λ

βl = 2π/λ × l

l = 2 cm = 0.02 m

βl = 2π/0.0625 × 0.02=2.01 rad = 115.3°

1 rad = 180/π degrees

$Z_L=load\ impedance=37.5+j75\\\\Z_o=characteristic impedance = 75\ ohm\\\\\tilde {Z_L}=Z_L/Z_o=37.5+j75/75=0.5+j$

$\tilde {Z_{in}}=\frac{\tilde {Z_{L}}+jtan\beta l}{1+j\tilde {Z_{L}}tan\beta l}=\frac{0.5+j+jtan(115.2)}{1+j(0.5+j)tan(115.2)}=0.253-j0.274\\ \\Z_{in}=Z_o\tilde {Z_{in}}=75(0.253-j0.274)=19-j20.5\\\\\Gamma_L=\frac{Z_L-Z_0}{Z_L+Z_o}=\frac{37.5+j75-75}{37.5+j75+75}=0.62\angle 83^o\\\\\Gamma_{in}=\frac{Z_{in}-Z_0}{Z_{in}+Z_o}=\frac{19-j20.5-75}{19-j20.5+75}=0.62\angle -147^o\\\\VSWR=\frac{1+\rho}{1-\rho} =\frac{1+0.62}{1-0.62}=4.26$

6 0

2 years ago

1. What is engineering? Based on your current perception, in a few sentences define the term “engineering.”

andreev551 [17]

THANKSTHANSKTHANKSTHANSK

THANKSTHANSK

6 0

3 years ago

Calculate the length of a metal cylinder while it is subjected to a tensile stress of 10,000 psi. You are given the following da

anastassius [24]

Answer:

length of cylinder can not calculated

Explanation:

given data

tensile stress = 10,000 psi

Original length = 1

Original cross-sectional area = 0.1 in²

Yield strength, σy = 9 ksi

Young’s modulus, E = 1000 ksi

solution

we can see that here that applied stress is greater than yield stress of material that is express

1000 ksi > 9 ksi

so here hooks law and strain relation is not working

so length of cylinder can not calculated

as stress applied 10000 psi

5 0

3 years ago