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Juliette [100K]
3 years ago
11

You could be sued if you injure someone while rescuing them if...

Engineering
2 answers:
Leviafan [203]3 years ago
4 0
It’s 4. One is accidental and can be the fault of the car dealership or maker, same with two. Three is their fault, and you could sue them.
navik [9.2K]3 years ago
3 0
I believe number 4 I could be wrong but I think it’s 4
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A renewable item is something that is capable of being replaced naturally.
djverab [1.8K]
The answer is False.
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3 years ago
Read 2 more answers
In some companies, workers who increase the quantity or quality of their work receive a. benefits. c. performance bonuses. b. pe
Greeley [361]

Answer:

performance bonuses

Explanation:

3 0
3 years ago
Can the MOXIE created by NASA be used on earth
Alisiya [41]

Answer:

MOXIE is designed to generate up to 10 grams of oxygen per hour. This technology demonstration was designed to ensure the instrument survived the launch from Earth, a nearly seven-month journey through deep space, and touchdown with Perseverance on Feb

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3 years ago
The fracture toughness of a stainless steel is 137 MPa*m12. What is the tensile impact load sustainable before fracture that a r
Charra [1.4K]

Answer:

7.7 kN

Explanation:

The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.

It can be expressed by using the formula:

K = \sigma Y \sqrt{\pi a}

where;

fracture toughness K = 137 MPam^{1/2}

geometry factor Y = 1

applied stress \sigma = ???

crack length a = 2mm = 0.002

∴

137 =\sigma \times 1  \sqrt{ \pi \times 0.002 }

137 =\sigma \times 0.07926

\dfrac{137}{0.07926} =\sigma

\sigma = 1728.489 MPa

Now, the tensile impact obtained is:

\sigma = \dfrac{P}{A}

P = A × σ

P = 1728.289 × 4.5

P = 7777.30 N

P = 7.7 kN

7 0
3 years ago
An aluminum oxide component must not fail when a tensile stress of 12.5 MPa is applied. Determine the maximum allowable surface
aivan3 [116]

Answer:

1.44 mm

Explanation:

Compute the maximum allowable surface crack length using

C=\frac {2E\gamma}{\pi \sigma_c^{2}} where E is the modulus  of elasticity, \gamma is surface energy and \sigma_c is tensile stress

Substituting the given values

C=\frac {2\times 393\times 10^{9}\times 0.9}{\pi\times (16\times 10^{6})^{2}= 0.001441103 m\approx 1.44mm

The maximum allowable surface crack is 1.44 mm

4 0
3 years ago
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