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Serga [27]
3 years ago
9

A substance that produces hydrogen ions in solution is

Physics
2 answers:
jeka943 years ago
7 0

Answer:

vejebrv

Explanation:

what is the name in fortnite for the best first to get xp ijeie3j2bneyeeg4ggb4hbb44bb4nn3n3jekeieikkkkkijfurnrhrhhehnrrnjrhhhrri48488hrh4hh4hhe4h44h4h4hu44i4i

lisabon 2012 [21]3 years ago
3 0
Answer- Acid


Explanation- an acid is a substance that contains hydrogen and ionizes to produce hydrogen ions in aqueous solutions. A base is a substance that contains hydroxide in the formula and produces a hydroxide ions in solution.
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Which imaging technique uses x-rays to make cross-sectional images of the body? mri ct scan x-ray imaging ultrasound
Vitek1552 [10]

Computed tomography (CT) is an imaging tool that combines x-rays with computer technology to produce a more detailed, cross-sectional image of your body.

<h3>What is CT-Scane?</h3>

Computed tomography (CT) scan is a useful diagnostic tool for detecting diseases and injuries.

Computed tomography (CT) is an imaging tool that combines x-rays with computer technology to produce a more detailed, cross-sectional image of your body.

It uses a series of X-rays and a computer to produce a 3D image of soft tissues and bones.

CT is a painless, noninvasive way for your healthcare provider to diagnose conditions.

A CT scan lets your doctor see the size, shape, and position of structures that are deep inside your body, such as organs, tissues, or tumors.

Hence Computed tomography (CT) is an imaging tool that combines x-rays with computer technology to produce a more detailed, cross-sectional image of your body.

To know more about CT-scane follow

brainly.com/question/1524856

5 0
2 years ago
Answer True or Flase1-Electric potential due to a uniform E field doesn’t change with location.2-The equipotential surfaces asso
TEA [102]

Answer:

1. False

2. True

3. True

Explanation:

1- False —> The relation between electric potential and electric field is given such that

-\int\limits^a_b \vec{E}d\vec{l} = V_{ab}

Therefore, for a uniform E field, electric potential is linearly proportional to the distance.

2- True —> The electric field lines always cross the equipotential lines perpendicularly.

3- True —> In order to be a potential difference, one source of electric field is enough. The electric potential will decrease radially according to the following formula:

V = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}

There is no test charge in the formula, only the source charge. Even when there is no test charge, the potential difference between points in space can exist.

3 0
3 years ago
A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou
Hoochie [10]

Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.

The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

<u>Mathematical expression for the Newton's second law of motion is given as:</u>

F=\frac{dp}{dt} ............................................(1)

where:

dp = change in momentum

dt = time taken to change the momentum

We know, momentum:

p=m.v

Now, equation (1) becomes:

F=\frac{d(m.v)}{dt}

<em>∵mass is constant at speeds v << c (speed of light)</em>

\therefore F=m.\frac{dv}{dt}

and, \frac{dv}{dt} =a

where: a = acceleration

\Rightarrow F=m.a

also

F\propto \frac{1}{dt}

so, more the time, lesser the force.

<em>& </em><u><em>Impulse:</em></u>

I=F.dt

I=m.a.dt

I=m.\frac{dv}{dt}.dt

I=m.dv=dp

∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.

So, the impulse in both the cases will be same.

4 0
3 years ago
A ray of light passes from one material into a material with a higher index of refraction. Determine whether each of the followi
nikitadnepr [17]

Answer:

a) the Angle te also decreases , b) decrease, c) unchanged , d) the speed decrease , e) unchanged

Explanation:

When a ray of light passes from one transparent material to another, it must comply with the law of refraction

     n1 sin θ₁ = n2 sin θ₂

Where index v1 is for the incident ray and index 2 for the refracted ray

With this expression let's examine the questions

a) They indicate that the refractive index increases,

      sin θ₂ = n₁ / n₂ sin θ₁

     θ₂ = sin⁻¹ (n₁ /n₂   sin θ₁)

    As m is greater than n1 the quantity on the right is less than one, the whole quantity in parentheses decreases so the Angle te also decreases

Answer is decrease

b) The wave velocity eta related to the wavelength and frequency

      v = λ f

The frequency does not change since the passage from one medium to the other is a process of forced oscillation, resonance whereby the frequency in the two mediums is the same.

The speed decreases with the indicated refraction increases and therefore the wavelength decreases

      λ = λ₀ / n

The answer is decrease

c) from the previous analysis the frequency remains unchanged

d) the refractive index is defined by

       n = c / v

So if n increases, the speed must decrease

The answer is decrease

e) the energy of the photon is given by the Planck equation

      E = h f

Since the frequency does not change, the energy does not change either

Answer remains unchanged

7 0
3 years ago
Convert the number from scientific into standard notation: 5.9 x 10-2
guapka [62]
Move the decimal point to:
Left : (if the exponent of ten is a negative number -) ... OUR CASE HERE (-2)
or to
Right : (if the exponent is positive +).

You should move the point as many times as the exponent indicates.
Do not write the power of ten anymore.

So, standard form is:
Two points to the left {Exponent of Ten is Negative (-2)}
0.059 ... (without the 10)
6 0
3 years ago
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