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GarryVolchara [31]
2 years ago
9

Name two hormones produced by the endocrine system and how they work with other organs system to maintain homeostasis

Physics
1 answer:
Serga [27]2 years ago
7 0

Answer:

insulin and glucagon, to regulate blood-glucose levels

Explanation:

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Visible light with a wavelength of 480 nm appears
DedPeter [7]
Green would be best 
hope it help
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3 years ago
Focus groups are one of the most widely used ________ methods to gain greater understanding of a current problem or to develop p
Zigmanuir [339]

Answer:

Exploratory

Explanation:

<u>Focus groups</u>

It is a small group of 8-12 respondents guided by a moderator through a thorough debate on a specific subject or idea.It's great for generation of ideas, brainstorming, insight into motives, attitudes, and perceptions. it can  show likes, dislikes,emotional requirements and prejudices.

Exploratory methods are used to gain initial insights that could pave the way for further investigation.

Some of the exploratory methods are focus groups, Key informant,case studies,secondary data and observational data.

8 0
3 years ago
Which statement about moons is true?
kotegsom [21]

Answer:

A

Explanation:

All of the other answers don't make much sense

8 0
3 years ago
. A mass m is traveling at an initial speed of 25.0 m/s. It is brought to rest in a distance of 62.5 m by a net force of 15.0 N.
harkovskaia [24]

Answer:

m = 3 kg

The mass m is 3 kg

Explanation:

From the equations of motion;

s = 0.5(u+v)t

Making t thr subject of formula;

t = 2s/(u+v)

t = time taken

s = distance travelled during deceleration = 62.5 m

u = initial speed = 25 m/s

v = final velocity = 0

Substituting the given values;

t = (2×62.5)/(25+0)

t = 5

Since, t = 5 the acceleration during this period is;

acceleration a = ∆v/t = (v-u)/t

a = (25)/5

a = 5 m/s^2

Force F = mass × acceleration

F = ma

Making m the subject of formula;

m = F/a

net force F = 15.0N

Substituting the values

m = 15/5

m = 3 kg

The mass m is 3 kg

7 0
3 years ago
One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren
katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

\sigma = \frac{F}{A}

Where,

\sigma =Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

\sigma_y = \frac{F_y}{A}

where,

\sigma_y =Shear strength

F_y = Shear Force

A_0 =Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}

Resolving for F,

F= 1399.5N

PART B) We need here to apply the shear strength equation, then

\sigma_y = \frac{F_y}{A}

210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}

F_y = 945N

In such a way that the material is more resistant to tensile strength than shear force.

6 0
3 years ago
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