PHYSICS HIGH SCHOOL

Match these terms with the correct examples.1. liquid solution
2. gaseous solution
3. compound
4. colloid
5. element
a. oceans
b. clouds
c. oxygen
d. air
e. water

Answers

Answer 1

Answer:

1) Liquid solution: Oceans

2) Gaseous Solution: Air

3) Compound: Water

4) Colloid: Cloud

5) Element: Oxygen

Explanation:

Solutions are those mixtures which have uniform composition. The size of solute is around 1 Angstrom hence, the particles of solutes can not be seen and the interactions make the overall solution transparent.

Compounds are those substances which are formed chemically by the reaction of two different elements like,

O₂ + 2 H₂ → 2 H₂O

Colloids are also mixtures but these are intermediate between true solutions and suspensions. Also, the solute size is greater than that of solute size in solutions and is about 10 Angstrom. Examples include Clouds, Fog, Moist, Aerosol sprays e.t.c.

Elements are those substances which either are made up of single atoms like noble gases or are molecules made up of same elements like O₂, H₂, Cl₂, N₂ e.t.c.

Answer 2

Answer:

Use other awnser this one was wrong

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What did the protoplanets become?a. nebulae
b. planets
c. solar nebulae
d. planetesimals

Answers

What did the protoplanets become?

a. nebulae
b. planets
c. solar nebulae
d. planetesimals

The protoplanets become nebulae. The answer is letter A. The rest of the choices do not answer the question above.

The answer is B. Planets

A ball is thrown vertically upward with an initial velocity of 23.6 meters per second. What is the maximum height reached by the ball? Please include your answer with a unit symbol.

Answers

Answer:

Maximum height attained by the ball will be h = 28.416 m

Explanation:

We have given initial speed of the ball u = 23.6 m /sec

At maximum height velocity will be zero so final velocity v = 0 m/sec

Acceleration due to gravity $g=9.8m/sec^2$

From law of motion we know that $v^2=u^2-2gh$

$0^2=23.6^2-2* 9.8* h$

h = 28.416 m

So maximum height attained by the ball will be h = 28.416 m

How does electricity flow?A.)from low potential to high potential
B.)from conductor to insulator
C.)from high potential to low potential
D.)from series circuit to parallel circuit

Answers

The answer is C.) High to low is the same thing as positive to negative.
Hope this helps!

C.) electricity always flows from high to low

Calculate how much work is required to launch a spacecraft of mass m from the surface of the earth (mass mE, radius RE) and place it in a circular low earth orbit--that is, an orbit whose altitude above the earth's surface is much less than RE. (As an example, the International Space Station is in low earth orbit at an altitude of about 400 km, much less than RE

Answers

The work done to launch the spacecraft in low earth orbit is $(GM_emh)/(R_e(R_e+h))$

Work-energy theorem:

The gravitational force is a conservative force. So, the total energy of the system must be conserved.

According to the work-energy theorem:

work done = - change in potential energy of the system.

W = -ΔPE

Initially, the potential energy of the satellite on the surface is:

PE = $-(GM_em)/(R_e)$

where m is the mass of the satellite

Let the orbit be at a height of h from the surface, so the potential energy in the orbit is :

PE' = $-(GM_em)/(R_e+h)$

ΔPE = PE'-PE

ΔPE = $-GM_em((1)/(R_e+h)-(1)/(R_e))$

$\Delta PE=-(GM_emh)/(R_e(R_e+h))$

Now work done:

W = - ΔPE

Thus,

$W=(GM_emh)/(R_e(R_e+h))$

Learn more about work energy theorem:

brainly.com/question/13603991?referrer=searchResults

Answer:

Work done to shift the spacecraft from Earth surface to low Earth radius is given as

$W = (GM_e m)/(2R_e)$

Explanation:

As we know that spacecraft is at surface of Earth initially

So we will have

$U_i = -(GM_e m)/(R_e)$

now when it is at low radius Earth Orbit then we have

$U_f = -(GM_e m)/(2(R_e + h))$

now we know that work done to shift the spacecraft from Earth Surface to Low earth orbit is change in total energy

$W = -(GM_e m)/(2(R_e + h)) + (GM_e m)/(R_e)$

so we have

$W = (GM_e m)/(R_e) (-(1)/(2(1 + h/R_e)) + 1)$

$W = (GM_e m)/(R_e) (-(1)/(2)(1 - (h)/(R_e)) + 1)$

$W = (GM_e m)/(2R_e)(1 + (h)/(R_e))$

since we know h << Re

so work done is given as

$W = (GM_e m)/(2R_e)$

A mass that weighs 8 lb stretches a spring 24 in. The system is acted on by an external force of 4 sin 4t lb. If the mass is pulled down 6 in. and then released, determine the position of the mass at any time. Determine the first four times at which the velocity of the mass is zero

Answers

Answer:

$t = (\pi)/(8), (\pi)/(4), (3\pi)/(8), (3 \pi)/(4)$

Explanation:

The equation of force is

F = 4 sin 4 t

Compare with the standard equation

f = A sin wt

where, w is the angular frequency and A is the amplitude.

Now

w = 4 rad/s

Let the time period is T.

the relation for the time period is

$T = (2\pi)/(w)\n\nT = (2 \pi)/(4)\n\nT = (\pi)/(2)$

the time period is defined as the time taken by the body to complete one oscillation.

So, the velocity is zero at the extreme points where the object is at time, T/4 and its odd T/2, 3T/4, 3T/2, etc.

So, the velocity is zero at time

$t = (\pi)/(8), (\pi)/(4), (3\pi)/(8), (3 \pi)/(4)$

Final answer:

To determine the position of the mass and the times when the velocity is zero in a mass-spring system.

Explanation:

To determine the position of the mass at any time, we need to solve the equation of motion for the mass-spring system. The equation is:

mx'' + kx = Fext

where m is the mass, x is the displacement of the mass from its equilibrium position, k is the spring constant, and Fext is the external force. We can solve this differential equation to find the position of the mass as a function of time.

To determine the first four times at which the velocity of the mass is zero, we need to solve for the velocity of the mass, which is given by:

v = x'

We can find the times at which the velocity is zero by finding the values of t for which x' = 0.

Learn more about Mass-spring system here:

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The charge on an electron is:
positive
negative
neutral
unknown

Answers

negative

Firstly it is best to describe the shape and structure of an atom, an atom's system and structure is like of the solar planet, the heliocentric view where the sun is the neutrons and protons and the planets are the electrons. Moreover, the electrons are outside these neutrons and protons surrounding the entire system of its atomic structure. Its movement is like of a planet.

An energy level also known as electron shell, or just shell is a location around an atom where electrons with a specific energy value, or quantum number may be found.

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