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faust18 [17]
3 years ago
13

Assume that a team of engineers runs a friction test on the car and track. The team finds that the car loses 210 joules of energ

y over each meter of track. How much energy does the roller coaster lose from friction as it travels from the top of the first hill to the top of the second hill
Physics
2 answers:
Reika [66]3 years ago
0 0

The rate of energy loss from friction is 210 joules/meter of track. Multiply by the length of the track (320 meters) to find the total energy lost:

210 J/m × 320 m = 67,200 J

The roller coaster car loses 67,200 joules of energy as heat between the tops of the two hills.

diamong [38]3 years ago
0 0

Answer:

The rate of energy loss from friction is 210 joules/meter of track. Multiply by the length of the track (320 meters) to find the total energy lost:

210 J/m × 320 m = 67,200 J

The roller coaster car loses 67,200 joules of energy as heat between the tops of the two hills.

Explanation:

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A sample of metallic frewium weighs 185N on a spring scale in air. When immersed in pure water, the frewium pulls on the scale w
balu736 [363]

Wow !  This one could have some twists and turns in it.
Fasten your seat belt.  It's going to be a boompy ride.

-- The buoyant force is precisely the missing <em>30N</em> .

--  In order to calculate the density of the frewium sample, we need to know
its mass and its volume.  Then, density = mass/volume .

-- From the weight of the sample in air, we can closely calculate its mass.

   Weight = (mass) x (gravity)
   185N = (mass) x (9.81 m/s²)
   Mass = (185N) / (9.81 m/s²) = <u>18.858 kilograms of frewium</u> 

-- For its volume, we need to calculate the volume of the displaced water.

The buoyant force is equal to the weight of displaced water, and the
density of water is about 1 gram per cm³.  So the volume of the
displaced water (in cm³) is the same as the number of grams in it.

The weight of the displaced water is 30N, and weight = (mass) (gravity).

           30N = (mass of the displaced water) x (9.81 m/s²)

           Mass = (30N) / (9.81 m/s²) = 3.058 kilograms

           Volume of displaced water = <u>3,058 cm³</u>

Finally, density of the frewium sample = (mass)/(volume)

      Density = (18,858 grams) / (3,058 cm³) = <em>6.167 gm/cm³</em> (rounded)

================================================

I'm thinking that this must  be the hard way to do it,
because I noticed that

       (weight in air) / (buoyant force) =  185N / 30N = <u>6.1666...</u>

So apparently . . .

        (density of a sample) / (density of water) =

                                  (weight of the sample in air) / (buoyant force in water) .

I never knew that, but it's a good factoid to keep in my tool-box.


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3 years ago
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Answer:

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Explanation:

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What is the distance in m between lines on a diffraction grating that produces a second-order maximum for 775-nm red light at an
8090 [49]

Answer:

The distance is d =  1.747 *10^{-6} \ m  

Explanation:

From the question we are told that  

       The order of maximum diffraction is  m =  2

         The wavelength is   \lambda  =  775 nm  = 775 * 10^{-9}  \ m

         The angle is  \theta =  62.5^o

Generally the   condition for  constructive  interference for diffraction grating  is mathematically represented as

          dsin \theta  =  m *  \lambda

where  d is  the distance between the lines on a  diffraction grating

     So  

            d =  \frac{m *  \lambda  }{sin (\theta  )}

substituting values  

           d =  \frac{2  *  775 *1^{-9} }{sin ( 62.5  )}

          d =  1.747 *10^{-6} \ m

   

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