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scoray [572]
3 years ago
8

Compare and contrast the strength of the forces between two objects with a mass of 1 kg each, a charge of 10

Physics
1 answer:
DochEvi [55]3 years ago
3 0

Answer:

Let's see the similarities between the two forces

* are proportional to the product of a magnitude, mass or charge

* They are inversely proportional to the square of the distance

* They are long-range forces since zero is not made up to an infinite distance. The gravitational force is always attractive, the electrical force can be attractive or repulsive.

The differences in them

* The electric force in much greater than the gravitational force

* The gravitational force is always attractive, the electrical force can be attractive or repulsive.

Explanation:

Let's start by calculating each force.

Gravitational force

             F =G \frac{m_1m_2}{r^2}  

let's calculate

             F = 6.67 10⁻¹¹  1  1 / 1²

             F = 6.67 10⁻¹¹ N

Electric force

             F = k \frac{q_1q_2}{r^2}  

indicates that the charge is q = 10 C

            F = 9 10⁹ 10 10 / 1²

            F = 9 10¹¹ N

Let's see the similarities between the two forces

* are proportional to the product of a magnitude, mass or charge

* They are inversely proportional to the square of the distance

* They are long-range forces since zero is not made up to an infinite distance. The gravitational force is always attractive, the electrical force can be attractive or repulsive.

The differences in them

* The electric force in much greater than the gravitational force

* The gravitational force is always attractive, the electrical force can be attractive or repulsive.

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3 years ago
A child is riding a merry-go-round that is turning at 7.18 rpm. If the child is standing 4.65 m from the center of the merry-go-
iVinArrow [24]

Answer:

B) 3.50 m/s

Explanation:

The linear velocity in a circular motion is defined as:

v=\omega r(1)

The angular frequency (\omega) is defined as 2π times the frequency and r is the radius, that is, the distance from the center of the circular motion.

\omega=2\pi f(2)

Replacing (2) in (1):

v=2\pi fr

We have to convert the frequency to Hz:

7.18rpm*\frac{1Hz}{60rpm}=0.12Hz

Finally, we calculate how fast is the child moving:

v=2\pi(0.12Hz)(4.65m)\\v=3.5\frac{m}{s}

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The escape velocity on earth is 11.2 km/s. What fraction of the escape velocity is the rms speed of H2 at a temperature of 31.0
Sveta_85 [38]

To solve this problem it is necessary to apply the concept related to root mean square velocity, which can be expressed as

v_{rms} = \sqrt{\frac{3RT}{n}}

Where,

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Our values are given as

v_e =11.2km/s = 11200m/s

The temperature is

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Therefore the root mean square velocity would be

v_{rms} = \sqrt{\frac{3(8.314)(303)}{0.002}}

v_{rms} = 1943.9m/s

The fraction of velocity then can be calculated between the escape velocity and the root mean square velocity

\alpha = \frac{v_{rms}}{v_e}

\alpha = \frac{1943.9}{11200}

\alpha = 0.1736

Therefore the fraction of the scape velocity on the earth for molecula hydrogen is 0.1736

7 0
3 years ago
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