All categories

3 years ago

Describe the work performed by a ski lift in terms of kinetic and gravitational potential energy

2 answers:

7 0

Explanation : When the lift falls, then the gravitational energy changed into the kinetic energy. Gravitational potential is associated with the mass of the lift and its position. When lift starts from rest , the mechanical energy of the lift change in the form of potential energy.

When the lift down than the potential energy lost and kinetic energy is gained. Lift loses height and gains speed.

The work is performed by the lift, when lift covers the distance against the gravitational force.

Whitepunk [10]3 years ago

7 0

In order to understand how kinetic and potential energy works in an ski lift it is necessary to know its terms

Potential energy is the energy that a body located from its height above the ground has

Kinetic energy is the energy of a body that is in motion

for the ski lift:

Kinetic Energy: When the ski lift moves from floor to floor

Potential Energy: When the ski lift is standing on the floor and people enter.

Gravitational energy: the chairlift uses the force of gravity to lower and change from potential energy to kinetic energy, with its own weight

I hope its help you

You might be interested in

On a nice summer day,Kim takes her niece Madison for a walk in her stroller.If they start from rest and accelerate at a rate of

11111nata11111 [884]

2.5m/s

Explanation:

Given parameters:

Initial velocity = 0m/s

Acceleration = 0.5m/s²

time of travel = 5s

Solution:

Final velocity = ?

Solution:

Acceleration can be defined as the change in velocity with time:

Acceleration = $\frac{Final velocity - Initial velocity}{time}$

From the equation above, the unknown is final velocity:

Final velocity - initial velocity = Acceleration x time

since initial velocity = 0

Final velocity = 0.5 x 5 = 2.5m/s

Learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

6 0

3 years ago

A glass plate (n = 1.64) is covered with a thin, uniform layer of oil (n = 1.28). A light beam of variable wavelength from air i

Makovka662 [10]

Answer:

Explanation:

This is case of interference in thin films

for constructive interference in thin film the condition is

2μ t = (2n+1)λ/2 ; μ is refractive index of oil , t is thickness of oil , λ is wave length of light .

2 x 1.28 x t = λ/2 , if n = 0

2 x 1.28 x t = 605 /2

t = 118.16 nm .

the minimum non-zero thickness of the oil film required = 118.16 nm.

8 0

3 years ago

The periodic table arranges elements by increasing _________.

Olegator [25]

Answer:

Atomic number

Explanation:

Hope it helps you in your learning process.

4 0

3 years ago

Ron is on a Ferris wheel of radius 30 ft that turns counterclockwise at a rate of one revolution every 12 seconds. The lowest po

alexgriva [62]

Answer:

$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t + 45$

Explanation:

1 full revolution is $2\pi.$ let \theta be the angle of Ron's position.

At t = 0. $\theta = 0$

one full revolution occurs in 12 sec, so his angle at t time is

$\theta =2\pi \frac{t}{12} = \frac{\pi}{6}t$

r is radius of circle and it is given as

$x = rcos\theta$

$y = rsin\theta$

for r = 30 sec

$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t$

however, that is centered at (0,0) and the positioned at time t = 0 is (30,0). it is need to shift so that the start position is (30,45). it can be done by adding to y

$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t + 45$

7 0

3 years ago

Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 10

11111nata11111 [884]

Answer:

$1.62\times 10^{-8}\ \text{s}$

Explanation:

$\epsilon_0$ = Vacuum permittivity = $8.854\times 10^{-12}\ \text{F/m}$

$A$ = Area = $10\times 2\times 10^{-4}\ \text{m}^2$

$d$ = Distance between plates = 1 mm

$V_c$ = Changed voltage = 60 V

$V$ = Initial voltage = 100 V

$R$ = Resistance = $1000\ \Omega$

Capacitance is given by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.854\times 10^{-12}\times 10\times 2\times 10^{-4}}{1\times 10^{-3}}\\\Rightarrow C=1.7708\times 10^{-11}\ \text{F}$

We have the relation

$V_c=V(1-e^{-\dfrac{t}{CR}})\\\Rightarrow e^{-\dfrac{t}{CR}}=1-\dfrac{V_c}{V}\\\Rightarrow -\dfrac{t}{CR}=\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-CR\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-1.7708\times 10^{-11}\times 1000\ln(1-\dfrac{60}{100})\\\Rightarrow t=1.62\times 10^{-8}\ \text{s}$

The time taken for the potential difference to reach the required level is $1.62\times 10^{-8}\ \text{s}$ .

5 0

3 years ago