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balandron [24]
2 years ago
13

What are some contact forces in the geosphere?

Physics
1 answer:
aivan3 [116]2 years ago
4 0

1) Example of contact force: friction

2) Examples of non-contact forces: gravity and electromagnetic force

Explanation:

1)

Contact forces are forces that acts only when the objects involved are touching.

An example of contact force in the geosphere is friction. Friction is a force that acts when two objects slide past each other, and the surfaces of the two objects are in contact. Due to the presence of "microbumps" on the two surfaces, there is a resistive force opposing the motion of the two objects, and this force is called friction.

Friction also acts when an object is moving through a fluid, although it takes a different name: resistance. Also in this case, the resistance acts in the direction opposite to the motion of the object, slowing it down.

2)

Non-contact forces are forces that act from a distance, therefore they act even when the objects involved are not touching.

Examples of non-contact forces are:

  • Gravitational force: this is an attractive force that acts between any object with mass. Its magnitude is directly proportional to the product of the masses of the two objects and inversely proportional to the square of the distance between the two objects.
  • Electromagnetic force: this is a force exerted between electrically charged objects. Its magnitude is directly proportional to the product of the two charges and inversely proportional to the square of the distance between the charges. It can be attractive (if the charges have opposite sign) or repulsive (if the charges have same sign).

Learn more about forces:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

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8 0
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An RL circuit contains a resistor with R = 6800 Ω and an inductor with L = 2300 µH. If the impedance of this circuit is 160,000
Rainbow [258]

| Impedance | = √ [R² +(ωL)²]

R² = 6800² = 4.624 x 10⁷
 
(ωL)² = (2 · π · f · 2.3 · 10⁻³)²

          = 2.0884 x 10⁻⁴  f²

| Z | =  √[ (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²) ]  =  1.6 x 10⁵

     (1.6 x 10⁵)²  =  (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²)

     (2.56 x 10¹⁰) - (4.624 x 10⁷)  =  2.0884 x 10⁻⁴ f²


Frequency² =   (2.56 x 10¹⁰ - 4.624 x 10⁷)  /  2.0884 x 10⁻⁴

                    =       2.555 x 10¹⁰ / 2.0884 x 10⁻⁴

                    =          1.224 x 10¹⁴ 

                    =          122,400 GHz          <== my calculation

                                      11.1 MHz           <== online impedance calculator

Obviously, I must have picked up some rounding errors
in the course of my calculation. 
  











7 0
2 years ago
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