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4 years ago

8

There are many different ways to classify matter but one way to only use two categories for all matter. All matter can be classi

fied as either
A) solids or liquid

B) mixtures or substances

C) atoms or molecules

D) elements or compounds

2 answers:

Ronch [10]4 years ago

8 0

Answer:

b mixtures or substances

Explanation:

i seasrced

the web that is the anwer i got

marysya [2.9K]4 years ago

6 0

Answer:

the answer is b

Explanation:

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3) A tolley of mass 4kg, moving with a velocity

Leokris [45]

Answer:

3.57 m/s

Explanation:

The sum of the 2 momentums Is equal the finale momentums. so if momentums Is q, v Is velocity and m Is Mass, q3=m1*v1+m2**v2=16+9=25 m*kg/s

q3=m3*v3

v3=q3/m3=25/(4+3)=3.57m/s

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3 years ago

Why is air considered a fluid

34kurt

Answer:

A fluids is any substance that flows. Air is made of stuff, air particles, that are loosely held together in a gas form. Although liquids are the most commonly recognized fluids, gasses are also fluids. Since air is a gas, it flows and takes the form of its container.

Explanation:

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3 years ago

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Most metals are good electrical conductors because (5 points) they hold their electrons they have a large amount of free electro

Hunter-Best [27]

Answer:

Metals are good conductors of electricity <u>because they have free electrons</u> which help in the transfer of charge from one point to another.

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3 years ago

What are the three important safety factors for surgical technologists to consider when exposed to ionizing radiation?

Gemiola [76]

Answer:

Exposure time limitation, shielding and distance.

Explanation:

Limitation of exposure time, since the dose received is directly proportional to the exposure time, so that, at a shorter time, lower dose. For this reason, planning is suggested, to reduce time.

Use of shields. This allows a reduction in the dose received by the technician when filtered by the barrier (screen). There are two types of shields or screens, the primary barriers (attenuate the radiation of the primary beam) and the secondary barriers (avoid diffuse radiation).

Distance to the radioactive source. The dose received is inversely proportional to the square of the distance to the radioactive source. Therefore, if the distance is doubled, the dose received will decrease by a quarter. Reason for this, it is advisable to use devices or remote controls whenever possible.

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3 years ago

What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

Wittaler [7]

Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$

$v = \frac{2 \pi (1.97x10^{6} m)}{84600s}$

$v = 146.31m/s$

Finally, equation 1 can be used:

$a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}$

$a = 0.010m/s^{2}$

Hence, the centripetal acceleration is $0.010m/s^{2}$

To given the centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ it is gotten:

$\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}$

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3 years ago