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ANTONII [103]
2 years ago
12

What is the strength of an electric field that will balance the weight of a proton?

Physics
1 answer:
alukav5142 [94]2 years ago
3 0

The weight (force of gravity) on an object near the surface of the earth is given by

F = mg

where m is the object's mass and g is the acceleration of objects near the surface of the earth.

The electric force on a charged object is given by

F = Eq

where E is the electric field at the point where the object is located and q is the charge of the object.

The weight (force of gravity) of a proton will act downwards (toward the surface of the earth), so in order for this force to be balanced, there must be an electric field pointing upward such that the electric force is equal and opposite to the weight. Set the magnitudes of the weight and electric force equal to each other:

mg = Eq

E = mg/q

Known values:

m = 1.67×10⁻27kg (a quick Google search will yield the mass of a proton)

g = 9.81m/s²

q = 1.60×10⁻19C

Plug in the known values and solve for E:

E = 1.67×10⁻27×9.81/(1.6×10⁻19)

E = 1.02×10⁻⁷N/C

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4 0
2 years ago
Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ
bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; H_0 = 51km/s/Mly

Age of the universe; t = \ ?

We know that, the reciprocal of the Hubble's constant ( H_0 ) gives an estimate of the age of the universe ( t ). It is expressed as:

Age\ of\ Universe; t = \frac{1}{H_0}

Now,

Hubble's constant; H_0 = 51km/s/Mly

We know that;

1\ light\ years = 9.46*10^{15}m

so

1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m

Therefore;

H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

Now, we input this Hubble's constant value into our equation;

Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0
2 years ago
Which one of the following types of electromagnetic wave travels through space the fastest?
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Answer:

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Explanation:

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Answer

It will stay the same!

Explanation:

If you so happen to move something from left to right, the size of it is not being shrunk or expanded in any type of way, shape, or form.

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2 years ago
How do you express 78,000 in scientific notation?
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