What characteristic of life would maintaining this balance be

Answers

Answer 1

Answer:

Explanation:When any living organism gets thrown off balance, its body or cells help it return to normal. In other words, living organisms have the ability to keep a stable internal environment. Maintaining a balance inside the body or cells of organisms is known as homeostasis.

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Four small spheres, each of which you can regard as a point of mass 0.200 kg, are arranged in a square 0.400 m on a side and connected by light rods. Find the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane.

Answers

The moment of inertia of the system about an axis through the center of the square, perpendicular to its plane is $0.0636 \;\rm kg-m^(2)$ .

Given data:

The mass of each sphere is, $m = 0.200 \;\rm kg$ .

Length of side of square is, $L = 0.400 \;\rm m$ .

The expression for the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane is,

$I = 4 mR^(2)$

Here,

R is the distance between center of the square and the sphere. And its value is,

$R =(1)/(2)\sqrt{L^(2)+L^(2)}\nR =(1)/(2)\sqrt{0.400^(2)+0.400^(2)}\nR = 0.282 \;\rm m$

Then, moment of inertia is,

$I = 4 mR^(2)\nI = 4 * 0.200 * 0.282^(2)\nI = 0.0636 \;\rm kg-m^(2)$

Thus, the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane is $0.0636 \;\rm kg-m^(2)$ .

Learn more about moment of inertia here:

brainly.com/question/2176093?referrer=searchResults

The moment of inertia of the system about an axis through the center of the square, perpendicular to the plane is 0.064 kg.m²

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Further explanation

Let's recall Moment of Inertia formula as follows:

$\boxed{ I = m R^2 }$

where:

I = moment of inertia

m = mass of object

R = distance between the object and the axis of rotation.

Given:

mass of sphere = m = 0.200 kg

length of side = x = 0.400 m

Asked:

net moment of inertia = ΣI = ?

Solution:

Let's ilustrate this question as shown in the attachment.

Firstly , let's find distance between center of the square and the sphere:

$R = (1)/(2) √(x^2+x^2)$

$R = (1)/(2) √(2x^2)$

$R = (1)/(2)√(2) x$

$R = (1)/(2) √(2) (0.400)$

$\boxed{R = 0.200√(2) \texttt{ m}}$

$\texttt{ }$

Next , we could find total moment of inertia as follows:

$\Sigma I = mR^2 + mR^2 + mR^2 + mR^2$

$\Sigma I = 4mR^2$

$\Sigma I = 4(0.200)(0.200√(2))^2$

$\boxed{\Sigma I = 0.064 \texttt{ kgm}^2}$

$\texttt{ }$

Learn more

Ratio of Moment of Inertia : brainly.com/question/2176655
Net Torque on the objects : brainly.com/question/9612563
Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Rotational Dynamics

How many squares for three traits?

Answers

you just need one square but with 16 little boxes
list out all the possible combination for each set of traits
add then to the each side of the box and match them up

what happens to the period of revolution for the planets as they move farther away in position from the sun

Answers

It increases. Mercury takes 88 days to orbit the sun once. The Earth takes a year. Pluto takes 248 years.

Direct current is produced byan electric motor
an armature
a solenoid
a battery

Answers

the answer is a battery

The answer is a battery

Method and answer please.

Answers

Answer:

(a) 200.0628 N ; ( b) 4.08Kg

Explanation:

We know,

w = mg [here w= weight , m= mass , g= gravity]

now,

gravity on earth (g) =9.807 m/s²

mass of the spanner (m)=?

weight of the spanner on earth(w) = 40N

⇒ w = 40N

⇒m g =40N

⇒m = (40/g) Kg

⇒m= (40/9.807)Kg

∴m=4.08Kg

∴mass of the spanner on earth = 4.08Kg

We know,

Mass is the amount of matter in an object . Somass will remain same both on earth and jupiter.

So,

(b)The mass of the spanner on jupiter = 4.08Kg

Now,

The mass of the spanner on jupitar (m) = 4.08Kg

Gravity on jupiter (g*) = 5.g [5 times on earth]

= 5 . (9.807)m/s²

=49.035 m/s²

(a)

∴weight of the spanner on jupitar(w*) = m g*

=(4.08 x 49.035) N

=200.0628 N

Please answer as soon as possible.

A Physics question about electricity and circuits.

Answers

R=144 ohms with method shown on photo

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