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3 years ago

Describe the Milky Way. What does it contain?

1 answer:

8 0

It consists of our Solar system. Also, it contains many stars and alot of dust and gases. Our milky way is a galaxy full of amazing stuff, many stars blow up and cause auroras in the sky. If you go to watch the aurora borealis(northern lights) , its amazing and beautiful. The sky turns green and blue and the stars appear.

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Organic macromolecules called _______ are insoluble in water

NikAS [45]

Answer:

lipids are insoluble in water which is why lipids are often found in biological membranes and other waterproof coverings.

3 0

3 years ago

A 3000-N force gives an object an acceleration of 15 m/s2. The mass of the object is

7nadin3 [17]

<h3>Answer:</h3>

200 kg

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality<u> </u>

<u>Physics</u>

<u>Newton's Law of Motions </u>

Newton's 1st Law of Motion: An object at rest remains at rest and an object in motion stays in motion

Newton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration)

Newton's 3rd Law of Motion: For every action, there is an equal and opposite reaction<u> </u>

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] F = 3000 N

[Given] a = 15 m/s²

[Solve] m = <em>x</em> kg

<u>Step 2: Solve for </u><em><u>m</u></em>

Substitute in variables [Newton's Second Law of Motion]: 3000 N = m(15 m/s²)
[Mass] [Division Property of Equality] Isolate <em>m</em> [Cancel out units]: 200 kg = m
[Mass] Rewrite: m = 200 kg

5 0

2 years ago

Atoms in a stays in a fixed position I have no freedom to move

Triss [41]

Is a solid...
when atoms or molecules or particles are in a fixed position it is a solid, it only vibrates in its place.

8 0

3 years ago

A 75 kg baseball player runs at a velocity of 6 m/s before sliding to a stop at second base. a. What is the kinetic energy of th

lana [24]

Answer:

a. $\displaystyle k_o=1350\ J$

b. $\displaystyle k_1=0\ J$

c. $\Delta k=-1350\ J$

d. $W=-1350\ J$

e. $F=-675\ N$

Explanation:

<u>Work and Kinetic Energy </u>

When an object moves at a certain velocity v0 and changes it to v1, a change in its kinetic energy is achieved:

$\Delta k=k_1-k_0$

Knowing that

$\displaystyle k=\frac{mv^2}{2}$

We have

$\displaystyle \Delta k=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}$

The work done by the force who caused the change of velocity (acceleration) is

$\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}$

If we know the distance x traveled by the object, the work can also be calculated by

$W=F.x$

Being F the force responsible for the change of velocity

The 75 kg baseball player has an initial velocity of 6 m/s, then he slides and stops

a. Before the slide, his initial kinetic energy is

$\displaystyle k_o=\frac{mv_0^2}{2}$

$\displaystyle k_o=\frac{(75)6^2}{2}$

$\boxed{\displaystyle k_o=1350\ J}$

b. Once he reaches the base, the player is at rest, thus his final kinetic energy is

$\displaystyle k_1=\frac{(75)0^2}{2}$

$\boxed{\displaystyle k_1=0\ J}$

c. The change of kinetic energy is

$\Delta k=k_1-k_0=0\ J-1350\ J$

$\boxed{\Delta k=-1350\ J}$

d. The work done by friction to stop the player is

$W=\Delta k=k_1-k_0$

$\boxed{W=-1350\ J}$

e. We compute the force of friction by using

$W=F.x$

and solving for x

$\displaystyle F=\frac{W}{x}$

$\displaystyle F=\frac{-1350\ J}{2\ m}$

$\boxed{F=-675\ N}$

The negative sign indicates the force is against movement

6 0

3 years ago

A charge Q is to be divided into two parts, labeled 'q' and 'Q-q? What is the relationship of q to Q if, at any given distance,

Lisa [10]

Answer:

Explanation:

The two charges are q and Q - q. Let the distance between them is r

Use the formula for coulomb's law for the force between the two charges

$F = \frac{Kq_{1}q_{2}}{r^{2}}$

So, the force between the charges q and Q - q is given by

$F = \frac{K\left ( Q-q \right )q}}{r^{2}}$

For maxima and minima, differentiate the force with respect to q.

$\frac{dF}{dq} = \frac{k}{r^{2}}\times \left ( Q - 2q \right )$

For maxima and minima, the value of dF/dq = 0

So, we get

q = Q /2

Now $\frac{d^{2}F}{dq^{2}} = \frac{-2k}{r^{2}}$

the double derivate is negative, so the force is maxima when q = Q / 2 .

6 0

3 years ago