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2 years ago

5

The Moon is much smaller than Earth but is still made of rock that you could walk on. However, you would need a spacesuit to do

this. Why

1 answer:

weqwewe [10]2 years ago

4 0

Because there is no oxygen in space and we need oxygen to function so we need the suit to incapsulate us in oxygen so we can respire and so can our skin

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______ is the percent ionization of a weak acid in water increases as the concentration of acid decreases.

Nuetrik [128]

The ratio of concentration of ionized acid to the initial concentration of acid multiplied by 100 will give the percent ionization of a weak acid in water increases as the concentration of acid decreases.

Explanation:

Percent ionization is used for quantifying the number of ions present in the weak acid when dissolved in a solution. So it is similar to the pKa value. The percent ionization value can be determined as negative log of dissociation constant. Also the as the number of ions increases in weak acid, the concentration of acid will be decreasing . It can be calculated using the formula for percent ionization as follows:

$Percent ionization = \frac{Ionized acid concentration}{Initial concentration of acid} * 100$

As the water volume or concentration increases, the acid will get diluted much more thus leading to decrease in the concentration of acid.

So the ratio of concentration of ionized acid to the initial concentration of acid multiplied by 100 will give the percent ionization of a weak acid in water increases as the concentration of acid decreases.

7 0

3 years ago

A. Describe one meal that you typically order when eating out.

marta [7]

A. “I usually get 10 piece nugget meal with medium fries at McDonald’s with a McFlurry

B. I could replace the 10 piece nuggets with a grilled chicken wrap and replace the medium fries with a small fries then replace the mcflurry with either water or unsweetened tea.

3 0

3 years ago

Read 2 more answers

you should begin viewing a bacteria specimen with what objective lens? view available hint(s)for part g you should begin viewing

Sergio039 [100]

Some people view bacteria specimens with a 100x objective lens in order to see the smallest details.
Others may use a 10x objective lens for more general purposes, such as examining stained slides or pictures.
And still others may use a 40x objective lens to gain maximum resolution when viewing images of thick samples.

It is important to choose the appropriate magnification for your needs so that you can properly examine the specimen under study.

<h3>Why is the 100x objective lens necessary to see bacteria?</h3>

Bacteria must, of course, be viewed at the maximum magnification and resolution possible because to their small size.
Due to optical restrictions, this is approximately 1000x in a light microscope.
To improve resolution, the oil immersion method is performed. This calls for a unique 100x objective.

To learn more about bacterial specimen, visit:

brainly.com/question/1412064

#SPJ4

5 0

2 years ago

What is the volume of this bubble when it reaches the surface?

steposvetlana [31]

Answer:

Volume will be 15 mL. Solution:- If we look at the given information then it is Boyle's law as the temperature is constant and the volume changes inversely as the pressure changes. So, the volume of the air bubble at the surface will be 15 mL.

8 0

3 years ago

Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

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

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0

3 years ago