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Harman [31]
3 years ago
6

Arm coverings can NOT protect you from cuts or puncture wounds. Is this statement TRUE or FALSE?

Engineering
1 answer:
Alexus [3.1K]3 years ago
5 0

Answer:

true

Explanation:

it uses a period , not a question mark .

You might be interested in
A loss in value caused by an undesirable or hazardous influence offsite is which type of depreciation?
Lubov Fominskaja [6]

External depreciation may be defined as a loss in value caused by an undesirable or hazardous influence offsite.

<h3>What is depreciation?</h3>

Depreciation may be defined as a situation when the financial value of an acquisition declines over time due to exploitation, fray, and incision, or obsolescence.

External depreciation may also be referred to as "economic obsolescence". It causes a negative influence on the financial value gradually.

Therefore, it is well described above.

To learn more about Depreciation, refer to the link:

brainly.com/question/1203926

#SPJ1

5 0
2 years ago
A bar of 75 mm diameter is reduced to 73mm by a cutting tool while cutting orthogonally. If the mean length of the cut chip is 7
barxatty [35]

Answer:

r=0.31

Ф=18.03°

Explanation:

Given that

Diameter of bar before cutting = 75 mm

Diameter of bar after cutting = 73 mm

Mean diameter of bar d= (75+73)/2=74 mm

Mean length of uncut chip = πd

Mean length of uncut chip = π x 74 =232.45 mm

So cutting ratio r

Cutting\ ratio=\dfrac{Mean\ length\ of cut\ chip}{Mean\ length\ of uncut\ chip}

r=\dfrac{73.5}{232.45}

  r=0.31

So the cutting ratio is 0.31.

As we know that shear angle given as

tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }

Now by putting the values

tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }

tan\phi =\dfrac{0.31cos15 }{1-0.31sin15 }\

  Ф=18.03°

So the shear angle is 18.03°.

4 0
4 years ago
Which situation might cause potential hazards at a construction site?
nata0808 [166]

Answer:

D. protruding steel rebars .. #answerwithquality #BAL

5 0
3 years ago
Read 2 more answers
A race car is travelling around a circular track. The velocity of the car is 20 m/s, the radius of the track is 300 m, and the m
Zielflug [23.3K]

Answer:

μ = 0.136

Explanation:

given,

velocity of the car = 20 m/s

radius of the track = 300 m

mass of the car = 2000 kg

centrifugal force

F_c = \dfrac{mv^2}{r}

F_c = \dfrac{2000\times 20^2}{300}

F c = 2666. 67 N

F f= μ N

F f = μ m g

2666.67  =  μ × 2000 × 9.8

μ = 0.136

so, the minimum coefficient of friction between road surface and car tyre is equal to μ = 0.136

5 0
4 years ago
2. Consider Dekker’s algorithm written for an arbitrary number of processes by changing the statement executed when leaving the
neonofarm [45]

Answer:

Algorith does not work.

Explanation:

One of the ways to obtain the Dekker Algorithm is through a change in the declaration, that is, a declaration that can be executed at the exact moment it leaves the critical section. This way it is possible that the statement,

turn = 1-i / * P0 sets turn to 1 and P1 sets turn 0 * /

It can be changed to,

turn = (turn +1) \% n / * n = number or processes * /

The result will allow to define if it works or not, that is, if it is greater than 2 the algorithm will not be able to work.

Given this consideration we can say that,

<em>- The dead lock does not occur, because the mutual is imposed (if a resource unit has been assigned to a process, then no other process can access that resource).</em>

<em>- There is the possibility of starving if the shift is established in a non-contentious process.</em>

Directly it can be concluded that there is a possibility of starvation so the algorithm could not work, despite the fact that mutual exclusion guarantees that a dead block does not occur.

4 0
3 years ago
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