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3 years ago

9

A vertical spring has a mass hanging from it, which is displaced from the equilibrium position and begins to oscillate. At what

point does the system have the least potential energy?

1 answer:

Orlov [11]3 years ago

5 0

Answer:

the object has least potential energy at mean position of the SHM

Explanation:

If a block is connected with a spring and there is no resistive force on the system

In this case the total energy of the system is always conserved and it will change from one form to another form

So here we will say that

Kinetic energy + Potential energy = Total Mechanical energy

As we can say that total energy is conserved so here we have least potential energy when the system has maximum kinetic energy

So here we also know that at mean position of the SHM the system has maximum speed and hence maximum kinetic energy.

So the object has least potential energy at mean position of the SHM

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A student pushes on a crate with a force of 100 Newtons directed to the right.

romanna [79]

Answer:

100N

Explanation:

Newton's third law of motion

For every action, there is an equal and opposite reaction.

Therefore 100N of force is exerted by the crate on student as a reaction to his action

6 0

3 years ago

Explain the costruction and working of windmill

fiasKO [112]

A windmill produces wind without electricity. wind blows on a windmill and it and that’s how it produces wind. You can put a windmill in anywhere outdoors (ex. Deserts or plains)
—————————————————————

Hope this helped you:)

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3 years ago

A 460 W heating unit is designed to operate with an applied potential difference of 120 V (a) By what percentage will its heat o

dybincka [34]

Answer:

(a) = -0.16%

(b) = smaller

Explanation:

given

power = 460 W

potential difference = 120 V

(a) what percentage will its heat output drop if the applied potential difference drops to 110 V ?

we know $p = \frac{v^2}{R}$ .....................(i)

we need to find change in power

$\Delta P = \frac{\Delta (V^2)}{R}$

$\Delta P = \frac{2 V \Delta V}{R}$ ..............(ii)

from equations we get

$\frac{\Delta P}{P} = \frac{2 \Delta V}{V}$

$\frac{\Delta P}{P} = 2 \frac{110 -120}{120}$

$\frac{\Delta P}{P} = -2(\frac{10}{120})$

$\frac{\Delta P}{P} = - 0.16 %$

(b)

if we increase temperature resistance will increase and decrease with decrease in temperature and we know power is inversely proportional to resistance so if potential decrease and it would cause drop in power

and due to this increment of heating power resistance will decrease so actual drop in the power would be smaller

7 0

3 years ago

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is:

$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

$W_{loss} = 6574.75\,J$

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0

3 years ago

Which color has the lowest frequency?

Svet_ta [14]

<h2>RED!</h2><h3></h3><h3>On the visible spectrum, red has the lowest frequency.</h3><h3>(I'm an amateur astronomer, so I would know.)</h3>

4 0

3 years ago