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statuscvo [17]
3 years ago
10

The mathematical relationship between

Engineering
1 answer:
11Alexandr11 [23.1K]3 years ago
7 0
4) Ohms law thats the answer
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A. 50
Mademuasel [1]

Answer:ahahahahahah

Explanation:boyboy

6 0
3 years ago
A 1-ft rod with a diameter of 0.5 in. is subjected to a tensile force of 1,300 lb and has an elongation of 0.009 in. The modulus
iragen [17]

Answer:

E = 8.83 kips

Explanation:

First, we determine the stress on the rod:

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

where,

σ = stress = ?

F = Force Applied = 1300 lb

A = Cross-sectional Area of rod = 0.5\pi \frac{d^2}{4} = \pi \frac{(0.5\ in)^2}{4} = 0.1963\ in^2

Therefore,

\sigma = \frac{1300\ lb}{0.1963\ in^2} \\\\\sigma = 6.62\ kips

Now, we determine the strain:

strain = \epsilon = \frac{elongation}{original\ length} \\\\\epsilon = \frac{0.009\ in}{12\ in}\\\\\epsilon =  7.5\ x\ 10^{-4}

Now, the modulus of elasticity (E) is given as:

E = \frac{\sigma}{\epsilon}\\\\E = \frac{6.62\ kips}{7.5\ x\ 10^{-4}}

<u>E = 8.83 kips</u>

7 0
3 years ago
Two players find themselves in a legal battle over a patent. The patent is worth 20 for each player, so the winner would receive
Alborosie

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

5 0
3 years ago
A manometer consists of an inclined glass tube which communicates with a metal cylinder standing upright; liquid fills the appar
Vanyuwa [196]

Answer:

48.61

Explanation:

See attached diagram.

The level rise in the tube is l sin α.

The level drop in the cylinder (let's call it y) is:

π/4 D² y = π/4 d² l

D² y = d² l

y = l (d/D)²

The elevation difference is the sum:

l sin α + l (d/D)²

l (sin α + (d/D)²)

From Bernoulli's principle:

P = ρgl (sin α + (d/D)²)

Divide both sides by density of water (ρw) and gravity:

P/(ρw g) = (ρ/ρw) l (sin α + (d/D)²)

h = S l (sin α + (d/D)²)

If we disregard the level change in the cylinder:

h = S l (sin α)

We want the percent error between these two expressions for h to be 0.1% when α = 25°.

[ S l (sin α + (d/D)²) − S l (sin α) ] / [ S l (sin α + (d/D)²) ] = 0.001

[ S l sin α + S l (d/D)² − S l sin α ] / [ S l (sin α + (d/D)²) ] = 0.001

[ S l (d/D)² ] / [ S l (sin α + (d/D)²)]  = 0.001

(d/D)² / (sin α + (d/D)²) = 0.001

(d/D)² = 0.001 (sin α + (d/D)²)

(d/D)² = 0.001 sin α + 0.001 (d/D)²

0.999 (d/D)² = 0.001 sin α

d/D = √(0.001 sin α / 0.999)

When α = 25°:

d/D ≈ 0.02057

D/d ≈ 48.61

7 0
3 years ago
Examine the pressure-measuring device shown in the figure below. (a) What is the gauge pressure reading in psi at point A? (b) W
Hunter-Best [27]

Answer: 45

Explanation:just cuase I need to

4 0
3 years ago
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