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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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Answer:

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7 0
3 years ago
wo skaters collide and embrace in an inelastic collision. Alex's mass is 90 kg and his initial velocity is 1.5 m/s i . Barbara's
Natalija [7]

Answer:

The two skaters will move with a common speed of 1.19 m/s.

Explanation:

Given that,

Mass of Alex, m_1=90\ kg

Initial velocity of Alex, u_1=1.5i\ m/s

Mass of Barbara, m_2=57\ kg

Initial velocity of Barbara, u_2=2j\ m/s

After the collision, the two skaters move together at a common velocity. Let V is the common velocity. Using the conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V\\\\90\times 1.5i+57\times 2j=(90+57)V\\\\V=\dfrac{135i+114j}{147}\\\\V=\dfrac{135i}{147}+\dfrac{114j}{147}\\\\V=(0.91i+0.77j)\ m/s

Magnitude of final velocity:

|V|=\sqrt{0.91^2+0.77^2} \\\\|V|=1.19\ m/s

So, the two skaters will move with a common speed of 1.19 m/s.

3 0
4 years ago
Read 2 more answers
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