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2 years ago

9

Read the passage.

1 answer:

Archy [21]2 years ago

6 0

The claim being made in in the above passage is that " It makes financial sense to stop using the penny." (Option B)

<h3>What textual evidence backs up the above claim?</h3>

The textual evidence that supports the above claim is "Not only does it make financial sense to take the penny out of circulation, but it also makes environmental sense." [Para. 2]

Textual evidence is evidence related to a text which supports claims made in such a text.

Learn more about claims at:
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The penalty for littering 15 lb or less is _____.<br> A. $25<br> B. $50<br> C. $100<br> D. $150

Marina CMI [18]

Answer:

D the closest

Explanation:

i looked it up and 8t says 200

5 0

3 years ago

g A thin-walled pressure vessel 6-cm thick originally contained a small semicircular flaw (radius 0.50-cm) located at the inner

galben [10]

This question is not complete, the complete question is;

A thin-walled pressure vessel 6-cm thick originally contained a small semicircular flaw (radius 0.50-cm) located at the inner surface and oriented normal to the hoop stress direction. Repeated pressure cycling enabled the crack to grow larger. If the fracture toughness of the material is $88 Mpam^\frac{1}{2}$ , the yield strength equal to 1250 MPa, and the hoop stress equal to 300 MPa, would the vessel leak before it ruptured

Answer:

length of crack is 5.585 cm

we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures

Explanation:

Given the data in the question;

vessel thickness = 6 cm

fracture toughness k = $88 Mpam^\frac{1}{2}$

yield strength = 1250 MPa

hoop stress equal = 300 MPa

we know that, the relation between fracture toughness and crack length is expressed as;

k = (1.1)(2/π)(r√(πa))

where k is the fracture toughness, r is hoop stress and a is length of crack

so we rearrange to find length of crack

a = 1/π[( k / 1.1(r)(2/π)]²

a = 1/π[( kπ / 1.1(r)(2)]²

so we substitute

a = 1/π [( 88π / 1.1(300)(2/π)]²

a = 1/π[ 0.1754596 ]

a = 0.05585 m

a = 0.05585 × 100 cm

a = 5.585 cm

so, length of crack is 5.585 cm

we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures

8 0

3 years ago

A new approval process is being adapted by Ursa Major Solar. After an opportunity has been approved, the contract is sent to the

BigorU [14]

Install an app :] i think lol

4 0

3 years ago

Re armature of a 4 pole DC generator is required to generate an emf of 520v on open circuit when revolving at a speed of 660rpm.

bija089 [108]

Since the armature is wave wound, the magnetic flux per pole is 0.0274 Weber.

<u>Given the following data:</u>

Emf = 520 Volts

Speed = 660 r.p.m

Number of armature conductors = 144 slots

Number of poles = 4 poles

Number of parallel paths = 2

To find the magnetic flux per pole:

Mathematically, the emf generated by a DC generator is given by the formula;

$E = \frac{\theta ZN}{60}$ × $\frac{P}{A}$

<u>Where:</u>

E is the electromotive force in the DC generator.

Z is the total number of armature conductors.

N is the speed or armature rotation in r.p.m.

P is the number of poles.

A is the number of parallel paths in armature.

Ф is the magnetic flux.

First of all, we would determine the total number of armature conductors:

$Z = 144$ × $2$ × $3$

Z = 864

Substituting the given parameters into the formula, we have;

$520 = \frac{\theta (864)(660)}{60}$ × $\frac{4}{2}$

$520 = \theta (864)(11)$ × $2$

$520 = 19008 \theta \\\\\Theta = \frac{520}{19008}$

<em>Magnetic flux </em><em>=</em><em> 0.0274 Weber.</em>

Therefore, the magnetic flux per pole is 0.0274 Weber.

Read more: brainly.com/question/15449812?referrer=searchResults

5 0

2 years ago

Can you solve this question

Alecsey [184]

Answer:

eojcjksjsososisjsiisisiiaodbjspbcpjsphcpjajosjjs ahahhahahahahahahahahahahahahhhahahahaahahhahahahahaahahahahaha

6 0

3 years ago

Read 2 more answers