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3 years ago

A book weighing 2.0 Newtons is lifted 3.0 meters in 4.0 seconds. How much work was done? SHOW WORK

Physics

1 answer:

qaws [65]3 years ago

7 0

6 Ns is the work done on the book to lift it by 3 m.

Answer:

Explanation:

Work done is the measure of amount of force required to move an object from its initial position to any desired position. So mathematically it can be calculated as the product of force acting on that object to the displacement of that object. Thus, work done on any object or by any object is directly proportional to the force acting on that object and the displacement of that object.

Work done = Force × displacement

As here the force is said to be 2 N and the displacement of the book is said to be 3 m, the work done on the book is

Work done = 2×3 = 6 N.s.

The time has no role to do on the work done calculation to be performed on the object.

Thus, 6 Ns is the work done on the book to lift it by 3 m.

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An assumption central to all scientific inquiry is causality, which means that

zavuch27 [327]

<h2>Answer</h2>

Causality means reason behind the process

<h2>Explanation</h2>

In the scientific inquiries, scientists perform different activities with the help of previous knowledge and scientific ways to renew the knowledge or discover the new ideas. It is a common belief that in every process that is happening in nature have many functions and reasons behind it. Scientists are always in the struggle to find these reasons and their importance regarding their happenings. When the scientists are in hurry to know about the causality, it means they are helping the scientific knowledge with the discovery of new reasons for happens.

7 0

3 years ago

A skateboarder is skating back and forth on the halfpipe as seen below. As he skates his energy transforms from potential energy

egoroff_w [7]

Answer:

Friction and air resistance cause some of his kinetic energy to be “lost”. This makes him slow down.

Explanation:

The law of conservation of energy states that in absence of frictional forces, the mechanical energy of an object (given by the sum of its kinetic and potential energy) is conserved. In such a situation, the skateboarder would never stop his motion, because potential energy is continuously converted into kinetic energy and vice-versa, but the total energy remains the same so he would never stop.

In a real world, however, this is not true. In fact, in a real world some frictional force are present, in particular:

- friction: this force is due to the contact between the skateboard and the surface of the halfpipe, and its direction is always opposite to the motion of the skateboarder

- Air resistance: this force is due to the resistance opposed by the molecules of air that the skateboarder meets during his motion, and its direction is also opposite to the motion of the skateboarder

This two forces are said to be non-conservative forces, which means that they cause some of the mechanical energy of the skateboarder to be "lost", in the sense that it is dissipated as heat and it is no longer available for the skateboarder.

Therefore, the correct option is

Friction and air resistance cause some of his kinetic energy to be “lost”. This makes him slow down.

7 0

3 years ago

Two ice skaters stand in the middle of an ice rink. Drew has a mass of 75 kg, and Lily has a mass of 55 kg. Drew holds Lily, and

ss7ja [257]

PART a)

Before Drew throw Lily in forwards direction they both stays at rest

So initial speed of both of them is zero

So here we can say that initial momentum of both of them is zero

So total momentum of the system initially = ZERO

PART b)

Since there is no external force on the system of two

so there will be no change in the momentum of this system and it will remain same as initial momentum

So final momentum of both of them will be ZERO

PART c)

As we know that momentum of both will be zero always

so we have

$P_1 + P_2 = 0$

$75(v) + 55(2) = 0$

$v = 1.47 m/s$ in opposite direction

7 0

3 years ago

Read 2 more answers

Water (density = 1x10^3 kg/m^3) flows at 15.5 m/s through a pipe with radius 0.040 m. The pipe goes up to the second floor of th

RUDIKE [14]

Answer:

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

Explanation:

By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

$P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2}$ (1)

Where:

$P_{1}$ , $P_{2}$ - Pressures of the water on the first and second floors, measured in pascals.

$\rho$ - Density of water, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Speed of the water on the first and second floors, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the water on the first and second floors, measured in meters.

Now we clear the final speed of the water flow:

$\frac{\rho\cdot v_{2}^{2}}{2} = P_{1}-P_{2}+\rho \cdot \left[\frac{v_{1}^{2}}{2}+g\cdot (z_{1}-z_{2}) \right]$

$\rho\cdot v_{2}^{2} = 2\cdot (P_{1}-P_{2})+\rho\cdot [v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})]$

$v_{2}^{2}= \frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})$

$v_{2} = \sqrt{\frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2}) }$ (2)

If we know that $P_{1}-P_{2} = 0\,Pa$ , $\rho=1000\,\frac{kg}{m^{3}}$ , $v_{1} = 15.5\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $z_{1}-z_{2} = -3.5\,m$ , then the speed of the water flow in the pipe on the second floor is: