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3 years ago

What are some contact forces in the geosphere?

Physics

1 answer:

aivan3 [116]3 years ago

4 0

1) Example of contact force: friction

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

Explanation:

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.

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:

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Nataliya [291]

Answer:

The position of the particle is -2.34 m.

Explanation:

Hi there!

The equation of position of a particle moving in a straight line with constant acceleration is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the particle at a time t:

x0 = initial position.

v0 = initial velocity.

t = time

a = acceleration

We have the following information:

x0 = 0.270 m

v0 = 0.140 m/s

a = -0.320 m/s²

t = 4.50 s (In the question, where it says "4.50 m/s^2" it should say "4.50 s". I have looked on the web and have confirmed it).

Then, we have all the needed data to calculate the position of the particle:

x = x0 + v0 · t + 1/2 · a · t²

x = 0.270 m + 0.140 m/s · 4.50 s - 1/2 · 0.320 m/s² · (4.50 s)²

x = -2.34 m

The position of the particle is -2.34 m.

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3 years ago

Two bodies made of the same material have same external dimension and appearance, but one is compact solid and other is hollow.

Bumek [7]

Answer:

Different

Explanation:

The hollow one will expand even more making it have a larger volume then the solid one so they are different

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3 years ago

A ski lift carries people along a 220-meter cable up the side of a mountain. Riders are lifted a total of 110 meters in elevatio

jeka94

The ideal mechanical advantage (IMA) can be determined by the following equation:

IMA= Input distance/Output distance

The Input distance and Output distance are:

Input distance=220 meters

Output distance=110 meters

When you substitute in the equation of the ideal mechanical advantage (IMA), you obtain:

IMA= Input distance/Output distance

IMA= 220 meters/110 meters

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3 years ago

Two cars, initially at rest and 5 km apart at t=0 , simultaneously move toward each other. Car A travels at a constant speed of

Anastasy [175]

Answer:

Explanation:

<u>Constant Speed Motion</u>

An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.

Expressed in a simple equation, we have:

$\displaystyle v=\frac{d}{t}$

Where

v = Speed of the object

d = Distance traveled

t = Time taken to travel d.

From the equation above, we can solve for d:

d = v . t

And we can also solve it for t:

$\displaystyle t=\frac{d}{v}$

Two cars are initially separated by 5 km are approaching each other at relative speeds of 55 km/h and 12 km/h respectively. The total speed at which they are approaching is 55+12 = 67 km/h.

The time it will take for them to meet is:

$\displaystyle t=\frac{5}{67}$

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3 years ago

A 0.15 kg project hits the ground with a speed of 11 m/s. The project comes to rest in 0.015 seconds. What is the net force?

Blababa [14]

Answer:

Explanation:

ACCORDING TO NEWTONS SECOND LAW;

F = mass * acceleration

F = m(v-u/t)

m is the mass = 0.15kg

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u is the initial velocity = 0m/s

t is the time = 0.015

Substitute;

F = 0.15(11-0)/0.015

F = 0.15(11)/0.015

F = 1.65/0.015

F = 110N

Hence the net force is 110N

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2 years ago