Which of the following is an activity of daily living? jogging cleaning weightlifting all of the above

Lately, you have noticed some repetitive stress in your wrist. Which sign is most likely the cause of that stress and pain?

Cerrena [4.2K]

1, you might have been carrying things that are way too heavy for you.
2, you might have weak tendons.

3 0

3 years ago

This question allows you to practice proving a language is non-regular via the Pumping Lemma. Using the Pumping Lemma (Theorem 1

Ulleksa [173]

Answer:

L is not a regular language with formal proofs

Explanation:

(a) To prove that L is not a regular language, we will use a proof by contradiction. the assumption entails that L is a regular language. Then by the Pumping Lemma for Regular Languages, 

there exists a pumping length p for L such that for any string s ∈ L where |s| ≥ p, 

s = xyz subject to the following conditions: 

(a) |y| > 0 

(b) |xy| ≤ p, and 

(c) ∀i > 0, xyi 

z ∈ L

(b) To determine that L is not a regular language, we mke use of proof by contradiction. lets assume, that L is regular. Then by the Pumping Lemma for Regular Languages, it states also,

The pumping length, p for L such that for any string s ∈ L where |s| ≥ p, s = xyz subject to the condtions as follows : 

(a) |y| > 0 

(b) |xy| ≤ p, and 

(c) ∀i > 0, xyi 

z ∈ L. 

Choose s = 0p10p 

. Clearly, |s| ≥ p and s ∈ L. By condition (b) above, it follows is shown. by the first condition x and y are zeros.

for some k > 0. Per (c), we can take i = 0 and the resulting string will still be in L. Thus, xy0 

z should be in L. xy0 

z = xz = 0(p−k)10p 

It is shown that is is not in L. This is a contraption with the pumping lemma. our assumption that L is regular is incorrect, and L is not a regular language

6 0

3 years ago

Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm an

ser-zykov [4K]

Answer:

volumetric flow rate = $0.0251 m^3/s$

Velocity in pipe section 1 = $6.513m/s$

velocity in pipe section 2 = 12.79 m/s

Explanation:

We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.

The density of water is = 997 kg/m³

density = mass/ volume

since we are given the mass, therefore, the volume will be mass/density

25/997 = $0.0251 m^3/s$

volumetric flow rate = $0.0251 m^3/s$

Average velocity calculations:

Pipe section A:

cross-sectional area =

$\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2$