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3 years ago

What is the 4th state of matter and it’s properties?

2 answers:

5 0

The 4th state is plasma. It is <span>composed of the building blocks of all matter & is essentially electrically charged particles at high energy.</span>

ddd [48]3 years ago

3 0

The 4 states of matter are solid, gas liquid, and plasma.

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For elevator questions, does the weight of a person increase when the elevator is going up or if it's going down-

TiliK225 [7]

The weight of a person increase when the elevator is going up.

<h3>Weight of the person in the elevator</h3>

The weight of the person in the elevator is calculated as follows;

<h3>When the person is going up</h3>

F = ma + mg

F = m(a + g)

where;

a is acceleration of the person
g is acceleration due to gravity

<h3>When the person is going down</h3>

F = mg - ma

F = m(g - a)

Thus, the weight of a person increase when the elevator is going up.

Learn more about weight here: brainly.com/question/2337612

#SPJ1

6 0

2 years ago

For this problem, imagine that you are on a ship that is oscillating up and down on a rough sea. Assume for simplicity that this

ikadub [295]

Answer:

no idea

Explanation:

7 0

3 years ago

An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

3 years ago

While John is traveling along an interstate highway, he notices a 150 mi marker as he passes through town. Later John passes a 1

Lerok [7]

Answer:

Part a)

distance = 112 miles

Part b)

current position = 112 miles from the position of town

Explanation:

Part a)

Since the distance marker is showing the distance between the town and the position of john at all time

so here we have

$d = 112 miles$

Part b)

Current position of John is given as

$r = 112 miles$

from the position of the town

7 0

3 years ago

Read 2 more answers

Help mee! Physics i think!

xxTIMURxx [149]

Answer:

Use the form of equation:

Q=mL

You have the specific latent heat of vaporization L = 2.260*10^{6}

And Q, the heat energy supplied, which equals 1695 KJ = 1695*10^{3} J

So you can get the mass by substitution in the formula below.

5 0

3 years ago