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Virty [35]
2 years ago
14

What magnitude charge creates a 32.08 N/C electric field at a point 2.84 m away?

Physics
1 answer:
sammy [17]2 years ago
5 0

Answer:

Q = 29.4 x 10⁻⁹ C.

Explanation:

Electric field due to a charge Q at distance d is given by coulomb law as follows

Electric field

E = k Q /d²

where for air k which is a constant is 9 x 10⁹

E=\frac{k\times Q}{d^2}

Given E = 32.08 , d = 2.84 m

Putting these values in the relation above, we have

[tex]32.08=\frac{9\times 10^9\times Q}{(2.84)^2}[/tex]

Q = 29.4 x 10⁻⁹ C.

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3 years ago
Physical Science - 02.05 - Question #3
Reika [66]

Answer:

1. They should not trust this study. 2. Pseudoscience.

Explanation:

Because they drank the water and after 24 hours, they said that it was gone. Head aches go away sooner than that. Another reason is that some people could have lied about having a headache just so they could have the water or their headache could have gone away before they drank the water.

6 0
2 years ago
Calculate the speed of a proton after it accelerates from rest through a potential difference of 350 V.
AVprozaik [17]

The speed of a proton after it accelerates from rest through a potential difference of 350 V is 25.86 \times 10^4 ~m/s.

Initial velocity of the proton u = 0

Given potential difference \Delta V = 350V

let's assume that the speed of the proton is v,

Since the proton is accelerating through a potential difference, proton's potential energy will change with time. The potential energy of a particle of charge q when accelerated with a potential difference \Delta V is,

    U = q \Delta V

Due to Work-Energy Theorem and Conservation of Energy - <em>If there is no non-conservative force acting on a particle then loss in Potential energy  P.E must be equal to gain in Kinetic Energy K.E</em> i.e

\Delta K = \Delta V

If the initial and final velocity of the proton is u and v respectively then,

change in Kinetic Energy  \implies  \Delta K = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - 0

change in Potential Energy \implies \Delta U = q\Delta V

from conservation of energy,

             v= \sqrt{\frac{2q\Delta V}{m}}

so,         v = \sqrt{\frac{2\times 350 \times 1.6\times 10^{-19}}{1.67 \times 10^{-27}}

                = 25.86 \times 10^4 ~m/s

To read more about the conservation of energy, please go to brainly.com/question/14668053

7 0
1 year ago
For the potential , determine the number of bound states?A)6 b)4 c) 3 d) 1
s2008m [1.1K]

Do you see that blank, open space after the word "potential ..." ?

There's supposed to be a number there that actually tells us the value of the potential.  Without that number ... and a lot more description of the whole scenario here ... there's no possible answer to the question.

7 0
3 years ago
One mole of magnesium (6 × 1023 atoms) has a mass of 24 grams, as shown in the periodic table on the inside front cover of the t
natka813 [3]

This question involves the concepts of density, volume, and mass.

The approximate diameter of a magnesium atom is "3.55 x 10⁻¹⁰ m".

<h3>STEP 1 (FINDING MASS OF INDIVIDUAL ATOM)</h3>

It is given that:

Mass of one mole = 24 grams

Mass of 6 x 10²³ atoms = 24 grams

Mass of 1 atom = \frac{24\ grams}{6\ x\ 10^{23}\ atoms} = 4 x 10⁻²³ grams

<h3>STEP 2 (FINDING VOLUME OF A SINGLE ATOM)</h3>

\rho = \frac{m}{V}\\\\V=\frac{m}{\rho}

where,

  • \rho = density = 1.7 grams/cm³
  • m = mass of single atom = 4 x 10⁻²³ grams
  • V = volume of single atom = ?

Therefore,

V=\frac{4\ x\ 10^{-23}\ grams}{1.7\ grams/cm^3}

V = 2.35 x 10⁻²³ cm³

<h3>STEP 3 (FINDING DIAMETER OF ATOM)</h3>

The atom is in a spherical shape. Hence, its Volume can be given as follows:

V =\frac{\pi d^3}{6}\\\\d=\sqrt[3]{ \frac{6V}{\pi}}\\\\d=\sqrt[3]{ \frac{6(2.35\ x\ 10^{-23}\ cm^3)}{\pi}}

d = 0.355 x 10⁻⁷ cm = 3.55 x 10⁻¹⁰ m

Learn more about density here:

brainly.com/question/952755

7 0
2 years ago
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