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3 years ago

Can somebody help me with that

Engineering

1 answer:

skelet666 [1.2K]3 years ago

4 0

Answer:

I think it's 23 ohms.

Explanation:

Not entirely sure about it.

hope this helps

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A concentrated load P is applied to the upper end of a 1.47-m-long pipe. The outside diameter of the pipe is D = 112 mm and the

myrzilka [38]

Answer:

Pmax = 38251.73 N

Explanation:

Given info

L = 1.47 m

D = 112 mm ⇒ R = D/2 = 112/2 mm = 56 mm

d = 101 mm ⇒ r = D/2 = 101/2 mm = 50.5 mm

a) We can apply the following equation in order to get Q (First Moment of Area):

Q = 2*(A₁*y₁-A₂*y₂)

where

A₁ = π*R² = π*(56 mm)² = 3136 π mm²

y₁ = 4*R/(3*π) = 4*56/(3*π) mm = 224/(3*π) mm

A₂ = π*r² = π*(50.5 mm)² = 2550.25 π mm²

y₂ = 4*r/(3*π) = 4*50.5/(3*π) mm = 202/(3*π) mm

then

Q = 2*(3136 π mm²*224/(3*π) mm-2550.25 π mm²*202/(3*π) mm)

⇒ Q = 62437.833 mm³

b) If τallow = 83 MPa = 83 N/mm²

P = ?

We can use the equation

τ = V*Q / (t*I) ⇒ V = τ*t*I / Q

where

t = D - d = 112 mm - 101 mm = 11 mm

I = (π/64)*(D⁴-d⁴) = (π/64)*((112 mm)⁴-(101 mm)⁴) = 2615942.11 mm⁴

Q = 62437.833 mm³

we could also use this equation in order to get Q:

Q = (4/3)*(R³-r³)

⇒ Q = (4/3)*((56 mm)³-(50.5 mm)³) = 62437.833 mm³

then we have

V = (83 N/mm²)*(11 mm)*(2615942.11 mm⁴) / (62437.833 mm³)

⇒ V = 2942.255 N

Finally Pmax = V = 38251.73 N

6 0

3 years ago

Fill in the empty function so that it returns the sum of all the divisors of a number, without including it. A divisor is a numb

tekilochka [14]

Answer:

// Program is written in C++

// Comments are used to explain some lines

// Only the required function is written. The main method is excluded.

#include<bits/stdc++.h>

#include<iostream>

using namespace std;

int divSum(int num)

{

// The next line declares the final result of summation of divisors. The variable declared is also

//initialised to 0

int result = 0;

// find all numbers which divide 'num'

for (int i=2; i<=(num/2); i++)

{

// if 'i' is divisor of 'num'

if (num%i==0)

{

if (i==(num/i))

result += i; //add divisor to result

else

result += (i + num/i); //add divisor to result

}

cout<<result+1;

}

6 0

3 years ago

Q4. (20 points) For a bronze alloy, the stress at which plastic deformation begins is 271 MPa and the modulus of elasticity is 1

babunello [35]

Answer:

a) P = 86720 N

b) L = 131.2983 mm

Explanation:

σ = 271 MPa = 271*10⁶ Pa

E = 119 GPa = 119*10⁹ Pa

A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²

a) P = ?

We can apply the equation

σ = P / A ⇒ P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N

b) L₀ = 131 mm = 0.131 m

We can get ΔL applying the following formula (Hooke's Law):

ΔL = (P*L₀) / (A*E) ⇒ ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)

⇒ ΔL = 2.9832*10⁻⁴ m = 0.2983 mm

Finally we obtain

L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm

3 0

3 years ago

Maximum iorsional shear siress.? Select one: a)- occurs at the center of a shaft. b)- occurs at the outer surface of a shaft c)-

pickupchik [31]

Answer:

b). Occurs at the outer surface of the shaft

Explanation:

We know from shear stress and torque relationship, we know that

$\frac{T}{J}= \frac{\tau }{r}$

where, T = torque

J = polar moment of inertia of shaft

τ = torsional shear stress

r = raduis of the shaft

Therefore from the above relation we see that

$\tau = \frac{T.r}{J}$

Thus torsional shear stress, τ is directly proportional to the radius,r of the shaft.

When r= 0, then τ = 0

and when r = R , τ is maximum

Thus, torsional shear stress is maximum at the outer surface of the shaft.

4 0

3 years ago

6. What are the potential consequences if a business does not follow regulations? (1 point)

dimaraw [331]

Answer:

all of the above

Explanation:

6 0

3 years ago