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8090 [49]
4 years ago
14

A person struggles very hard to lift a large boulder. He puts in so much effort, he starts to sweat, his heart rate increases, a

nd he gets really tired. All of his effort is for nothing, the boulder does not move at all. In this situation, has work been done? Explain!
*Use the scientific definition of "work" to answer this question.
Physics
1 answer:
Lesechka [4]4 years ago
5 0
No, he did not perform any work. Work is when you’re using energy which results in a force. Even though he was tired and sweaty, he did not move the boulder. So therefore he did not perform any work.
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For the question below which equation should be used?
velikii [3]

Answer:

Equation 2

Explanation:

Initial velocity ( Vo ) = 20 m/s

Acceleration ( a ) = 1.1 m/s^2

Distance ( d ) = 55 m

Final velocity ( Vi ) can be found using equation 2

4 0
2 years ago
A charged comb picking up bits of paper is an example of electrical conduction.
Elina [12.6K]
Yes. As bits of paper comes in electric field of charged comb, it attracts paper towards itself by a force called it's 'Electric force" & that process of flow of electrons is called "Electrical Conduction"

In short, Your Answer would be "True"

Hope this helps!
4 0
3 years ago
What does it mean for forces to be in equilibrium?.
lapo4ka [179]

Explanation:

If the size and direction of the forces on the object are exactly balanced , then there is no net force acting on the object

8 0
3 years ago
A force does work on an object if a component of the force is
IrinaK [193]

Answer:

A force does work on an object if a component of the force is parallel to the displacement of the object.

Explanation:

Work, a measurement of energy is said to be done when a force applied to an object results in the movement of that object to a certain distance and direction. Force is the act of push or pulls occurs on an object as a result of the interaction between that object with another one and displacement is the distance and direction covered by that object as a result of the force applied on it.

The work done (W) by a constant force (F) is equal to the product of the force in the direction of displacement of the object and the distance (d) moved by the object i.e., W = F * d.

The angle between the displacement and the force is θ, then the work done, W = Fd cos θ  ........ (1)

Positive work - Force acts in the same direction with respect to the displacement of the object. Here, θ is zero, so cos θ i.e., cos 0 is 1. Therefore, from the equation (1), W = Fd (i.e., work done by the force is positive).

Negative work - Force acts in the opposite direction with respect to the displacement of the object.  Here, θ is 180°, so cos θ i.e., cos 180° is -1. Therefore, from the equation (1), W = -Fd (i.e., work done by the force is negative).

If a force is applied to an object and it does not move, then the work done is zero i.e., W = F * 0 = 0. Also, if the force and displacement are at right angle to each other, then θ is 90°. Therefore, from the equation (1), W = 0 since cos 90° is zero.

7 0
4 years ago
A 3-ton Toyota land cruiser travelled at a speed of 45m/s, collides with a 2- ton comfort taxi which was parked in front of the
Olin [163]

Answer:

v = 27 m/s

Explanation:

To find the speed of cars after the collision  you take into account the momentum conservation law. Total momentum of both cars before the collision must be equal to the total momentum of both cars after the collision.

After the collision both cars traveled together, then you have:

m_1v_1+m_2v_2=(m_1+m_2)v   (1)

m1: mass of the Toyota = 3-ton = 3000 kg

m2: mass of the taxi = 2-ton = 2000kg

v1: speed of the Toyota before the collision = 45m/s

v2: speed of the car before the collision = 0 m/s (it is  at rest)

v: speed of both cars after the collision = ?

You solve the equation (1) for v:

v=\frac{m_1v_1+m_2v_2}{m_1+m_2}

Next, you replace the values of the rest of the variables:

v=\frac{(3000kg)(45m/s)+0kgm/s}{3000kg+2000kg}=27\frac{m}{s}

hence, just after the collision both cars have a speed of 27m/s

7 0
3 years ago
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