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Kruka [31]
3 years ago
8

When a rubberband is stretched all the way back, it is an example of which type of energy?

Physics
2 answers:
Murljashka [212]3 years ago
8 0
Elastic potential energy. because elastic is were something is pulled and when let go it pulls its self back to the original position so when you pull the rubberband it builds up energy to pull its self back together
 
avanturin [10]3 years ago
5 0
The answer is C. elastic potential energy
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A Chef is looking for a new frying pan that will allow her to cook food quickly at temperatures above 500°C. Use the data provid
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Answer:

Explanation:

1st one

What is your evidence?

Very heavy professional or restaurant pans will have iron handles, while those for home use will be made of brass or stainless steel. All are perfectly safe for oven use.

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2 years ago
A baseball m=.34kg is spun vertically on a massless string of length l=.52m. the string can only support a tension of tmax=9.9n
larisa86 [58]
<span>4.5 m/s This is an exercise in centripetal force. The formula is F = mv^2/r where m = mass v = velocity r = radius Now to add a little extra twist to the fun, we're swinging in a vertical plane so gravity comes into effect. At the bottom of the swing, the force experienced is the F above plus the acceleration due to gravity, and at the top of the swing, the force experienced is the F above minus the acceleration due to gravity. I will assume you're capable of changing the velocity of the ball quickly so you don't break the string at the bottom of the loop. Let's determine the force we get from gravity. 0.34 kg * 9.8 m/s^2 = 3.332 kg m/s^2 = 3.332 N Since we're getting some help from gravity, the force that will break the string is 9.9 N + 3.332 N = 13.232 N Plug known values into formula. F = mv^2/r 13.232 kg m/s^2 = 0.34 kg V^2 / 0.52 m 6.88064 kg m^2/s^2 = 0.34 kg V^2 20.23717647 m^2/s^2 = V^2 4.498574938 m/s = V Rounding to 2 significant figures gives 4.5 m/s The actual obtainable velocity is likely to be much lower. You may handle 13.232 N at the top of the swing where gravity is helping to keep you from breaking the string, but at the bottom of the swing, you can only handle 6.568 N where gravity is working against you, making the string easier to break.</span>
7 0
3 years ago
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If the _____ work on a simple machine equals the output work, the machine is said to be 100% efficient.
omeli [17]
<h3>Answer:</h3>

Input work

<h3>Explanation:</h3>

Concept being tested: Efficiency of machines

Therefore we need to know what is the efficiency of a machine

  • Efficiency of a machine is the ratio of work output of machine to the work input expressed as a percentage.
  • Thus;

Efficiency = (Work output ÷ Work input) × 100%

  • Therefore, if the work input is equal to the work output then the efficiency of the machine will be 100%.
  • Most machines are not 100% efficient due to loss of energy in form of heat due to friction of the moving parts of the machine.
3 0
3 years ago
I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi
Masteriza [31]

Answer:

<h3>a.</h3>
  • After it has traveled through 1 cm : I(1 \ cm) = 0.5488 I_0
  • After it has traveled through 2 cm : I(2 \ cm) = 0.3012 I_0
<h3>b.</h3>
  • After it has traveled through 1 cm : od( 1\ cm) =  0.2606
  • After it has traveled through 2 cm :  od( 2\ cm) =  0.5211

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient \mu the formula is:

I(x) = I_0 e^{-\mu x}

where I is the intensity of the beam, I_0 is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = 0.5488 \ I_0

After travelling 2 cm:

I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = 0.3012 \ I_0

<h2>b</h2>

The optical density od is given by:

od(x) = - log_{10} ( \frac{I(x)}{I_0} ).

So, after travelling 1 cm:

od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )

od( 1\ cm) = - log_{10} ( 0.5488 )

od( 1\ cm) = - (  - 0.2606)

od( 1\ cm) =  0.2606

After travelling 2 cm:

od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )

od( 2\ cm) = - log_{10} ( 0.3012 )

od( 2\ cm) = - (  - 0.5211)

od( 2\ cm) =  0.5211

3 0
3 years ago
What does increasing the speed of an object do to its potential energy?
JulijaS [17]
It does not affect the objects potential energy.
7 0
3 years ago
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