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lidiya [134]
3 years ago
14

How much work do you do when you lift a 100 N child 0.5 m?

Physics
1 answer:
Papessa [141]3 years ago
8 0

Answer:

<h2>50 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 100 × 0.5

We have the final answer as

<h3>50 J</h3>

Hope this helps you

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Match each term to the best description
Ronch [10]

Answer:

Coefficient of friction =  Magnitude depends on the interacting materials

Friction = A force that acts parallel to the surface.

kinetic friction = A force that is constant regardless of the applied force

Normal = A force that acts perpendicular to the surface

Static friction = A force that increases as applied force increases up to some maximum value

Explanation:

Let's define all the forces, and then let's solve the problem.

Normal force:

When an object rests on some place (like a book on the table) the force that causes the book to not fall through the table is called the normal force, which is usually equal to the weight of the object and acts perpendicular to the surface where the object is resting.

Friction force.

When an object moves (or tries to move) parallel to a surface, such that the object is in contact with that surface, there appears a force that opposes to the movement (the force is parallel to the surface, and in the opposite direction to the movement).

And this force can be written as:

F = -N*μ

Where μ is the coefficient of friction and N is the normal force.

If the object is not moving yet (but there is applied a force that would move the object) the coefficient is called the coefficient of static friction which increases in a given range, until it can't keep increasing and the object starts to move, while if the object is moving, the coefficient is called the coefficient of kinetic friction and it is constant, where usually the first one is larger than the second, and these coefficients depend on both materials, the surface one and the object one.

Then we have two friction forces, one called the kinetic friction and the other called the static friction.

Then:

Coefficient of friction =  Magnitude depends on the interacting materials

Friction = A force that acts parallel to the surface.

kinetic friction = A force that is constant regardless of the applied force

Normal = A force that acts perpendicular to the surface

Static friction = A force that increases as applied force increases up to some maximum value

5 0
3 years ago
Which stars are most common in the galaxy? why don't we see many of them?
Alina [70]
Scientists think 20 out of the 30 stars nearest to Earth are red dwarfs; however, none can be seen wuth the naked eye. The closest star ti the sun, Proxina Centauri, is a Red Dwarf.
4 0
3 years ago
Match each example to its type of energy.
PtichkaEL [24]
Solar energy - A

nuclear energy - B

fossil fuel energy - C

wind energy - D

geothermal energy - E

6 0
4 years ago
A positively charged particle Q1 = +45 nC is held fixed at the origin. A second charge Q2 of mass m = 6.5 μg is floating a dista
fenix001 [56]

Answer:

Q₂ = (9.83 × 10⁻⁹) C = +9.83 nC

Explanation:

The force of attraction/repulsion between the two charges is equal to the force of gravity on the charge 2.

The force of gravity on the charge two = mg

= 6.5 μg × 9.8 m/s² = (6.37 × 10⁻⁵) N

Since the gravity force is directed downwards, the force on charge 2 due to charge 1 has to be directed upwards, hence it is a force of repulsion.

The magnitude of the force between the two charges is given according to the Coulomb's law.

F = kQ₁Q₂/r²

k = Coulomb's constant = (9.0 × 10⁹) Nm²/C²

Q₁ = 45 nC = (45 × 10⁻⁹) C

Q₂ = ?

r = 25 cm = 0.25 m

F = (6.37 × 10⁻⁵) N

(6.37 × 10⁻⁵) = [(9.0 × 10⁹) × (45 × 10⁻⁹) × Q₂] ÷ 0.25²

Q₂ = 0.00000000983 C = (9.83 × 10⁻⁹) C = 9.83 nC

Hope this Helps!!!

7 0
4 years ago
Who is the following list referring to? Successful state senator and governor of Georgia, Only Georgian to serve as president of
topjm [15]

Answer:

Jimmy Carter

Explanation:

4 0
3 years ago
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