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ololo11 [35]
3 years ago
6

While it’s impossible to design a perpetual motion machine, that is, a machine that keeps moving forever, come up with ways to k

eep some type of periodic motion going for a very long time. Explain the limits on perpetual motion. Discuss what slows down a machine and how you might minimize those effects for at least one specific form of periodic motion, such as the motion of a spring or pendulum, or circular motion. Come up with an inventive solution!
Physics
2 answers:
svetlana [45]3 years ago
5 0

Answer:

Explanation:

It would be hard so many factors play a role in this angle kinetic potential energy gravity positioning height weight etc.to keep something as described moving as long as possible you'd have to build it heavily enough to balance out the momentum at each climax of the swing to balance with the gravitational pull of the earth core.I'd use a material such as copper for the variable.Also the weight distribution on the variables base plays a role.Take all the into account and experiment for a while!maybe one day you could find the right proportions in every role for kinetic potential gravitational momentum and weight to keep it going forever.

TEA [102]3 years ago
3 0
It would be hard so many factors play a role in this angle kinetic potential energy gravity positioning height weight etc.to keep something as described moving as long as possible you'd have to build it heavily enough to balance out the momentum at each climax of the swing to balance with the gravitational pull of the earth core.I'd use a material such as copper for the variable.Also the weight distribution on the variables base plays a role.Take all the into account and experiment for a while!maybe one day you could find the right proportions in every role for kinetic potential gravitational momentum and weight to keep it going forever. As said alotta factors play a huge role in this experiment  <span />
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3 0
3 years ago
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A 2.0 kg sphere with a velocity of 6.0 m/s collides head-on and elastically with a stationary 10 kg sphere
dmitriy555 [2]

Question: A 2.0 kg sphere with a velocity of 6.0 m/s collides head-on and elastically with a stationary 10 kg sphere, What is thier velocities after collision.

Answer:

v = 6 m/s, v' = 0 m/s

Explanation:

From the question,

For Elastic collision,

mu+m'u' = mv+m'v'......................... Equation 1

Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, v = final veolocity of the first sphere, v' = final velocity of the second sphere.

Also,

The relative velocity before collision = relative velocity after collision

u-u' = v-v'............................ Equation 2

Given:  m = 2 kg, m' = 10 kg, u = 6 m/s, u' = 0 m/s

Substitute into equation 1 and 2

2(6)+10(0) = 2v+10v'

2v+10v' = 12.............. Equation 3

6-0 = v-v'

v-v' = 6 ................... Equation 4

Solve equation 3 and 4 simultaneously.

v = 6+v'............. Equation 5

Substitute equation 5 into equation 3

2(6+v')+10v' = 12

12+2v'+10v' = 12

12v' = 12-12

v' = 0/12

v' = 0 m/s.

Also substitute the value of v' into equation 5

v = 6+0

v = 6 m/s

5 0
3 years ago
I need help with this one 2 if you guys don’t mind
zloy xaker [14]
3 is 3.81 meters

4 is 0.47 liters

5 is 4 cm

6 is 23 mm

7 is 53 m

8 is 1800 mg

9 is 31.07 mi

Hope I’m helping ya

6 0
4 years ago
Can you respond this two questions, please? :
Andrews [41]

Where's the diagram for question 1?

6 0
4 years ago
A 3.50 cm tall object is held 24.8 cm from a lens of focal length 16.0 cm. What is the image height?
Mandarinka [93]

Answer:

<h2>6.36 cm</h2>

Explanation:

Using the formula to first get the image distance

1/f = 1/u+1/v

f = focal length of the lens

u = object distance

v = image distance

Given f = 16.0 cm, u = 24.8 cm

1/v = 1/16 - 1/24.8

1/v = 0.0625-0.04032

1/v = 0.02218

v = 1/0.02218

v = 45.09 cm

To get the image height, we will us the magnification formula.

Mag = v/u = Hi/H

Hi = image height = ?

H = object height = 3.50 cm

45.09/24.8 = Hi/3.50

Hi = (45.09*3.50)/24.8

Hi = 6.36 cm

The image height is 6.36 cm

6 0
3 years ago
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