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Contact [7]
2 years ago
8

15 Points , Physics HW, can someone please show me step by step how to calculate percent difference for this problem. My numbers

are listed in the image. Thank you so much!

Physics
1 answer:
timofeeve [1]2 years ago
3 0

The percent difference between two numbers x and y is given by

\dfrac{|y-x|}x \times 100\%

The absolute value is there because we only care about the absolute percent difference, and not taking into account whether we go from x to y or vice versa. If we remove them, we have two possible interpretations of percent difference.

For example, the (absolute) percent difference between 3 and 6 is

\dfrac{|6-3|}3 \times 100\% = 100\%

In other words, we add 100% of 3 to 3 to end up with 6. This is the same as the percent difference going from 3 to 6. On the other hand, the percent difference going from 6 to 3 is

\dfrac{3-6}3\times100\%=-50\%

which is to say, we take away 50% of 6 away from 6 to end up with 3.

"Make comparisons to object measurements" tells us that the differences should be computed relative to "measurements for object". In other words, take x from the left column and y from the right column.

\dfrac{|7.1-7.3|}{7.1} \times 100\% \approx 2.82\%

\dfrac{|4.8-5.0|}{4.8} \times 100\% \approx 4.17\%

\dfrac{|7.2-7.5|}{7.2} \times 100\% \approx 4.17\%

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A student lifts a physical science book off the table and above their head. Is there work being done?
DiKsa [7]

Answer: A force must cause a displacement in order for work to be done. A book falls off a table and free falls to the ground. Yes. This is an example of work.

Explanation:

7 0
2 years ago
200 kW of solar radiation is shining on a 300 m^2 parking lot. What is the insulation on the parking lot?
ale4655 [162]

That's "<em><u>insolation</u></em>" ... not "insulation".

'Insolation' is simply the intensity of solar radiation over some area.

If 200 kW of radiation is shining on 300 m² of area, then the insolation is

           (200 kW) / (300 m²) = <em>(666 and 2/3) watt/m²</em> .

Note that this is the intensity of the <em><u>incident</u></em> radiation.  It doesn't say anything
about how much soaks in or how much bounces off.

Wait ! 
I just looked back at the choices, and realized that I didn't answer the question
at all.  I have no idea what  "1 sun"  means.  Forgive me.  I have stolen your
points, and I am filled with remorse.

Wait again !
I found it, through literally several seconds of online research.

           1 sun = 1 kW/m².

So 2/3 of a kW per m²  =  2/3 of 1 sun

That's between 0.5 sun and 1.0 sun.

I feel better now, and plus, I learned something.


7 0
3 years ago
Sitting in a chairlift, Rebecca has a gravitational potential energy of 5,997.6
stira [4]

Answer:

B) 12 m

Explanation:

Gravitational potential energy is:

PE = mgh

Given PE = 5997.6 J, and m = 51 kg:

5997.6 J = (51 kg) (9.8 m/s²) h

h = 12 m

8 0
3 years ago
If you slosh the water back and forth in a bathtub at the correct frequency, the water rises first at one end and then at the ot
notka56 [123]

Answer:

Velocity(v) = frequency(f) × wavelength

f = 0.3165

Wavelength = 2×length(L)

L = 157cm

Convert the length in centimetres to metre = 1.57m

v = 2×1.57 × 0.3165

v = 0.99m/s

Approx. 1m/s

Explanation:

The velocity of a wave is the product of its frequency and it's wavelength. The frequency is already known. The wavelength is the distance between two successive wave crests which is formed by sloshing water back and forth in the bath tub. Sloshing water to one end of the tub will produce a wave crest first at that end then the other completing a cycle. The wavelength will be twice the length of the bath tub as it is the distance that both crests are formed.

Wave crest is the highest point of a wave, and in this case is where the water rises to a high point in the bath tub

7 0
3 years ago
A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.
kap26 [50]

Answer:

The  frequencies are (f, f_1) =  (6615.4 \ Hz , 19846.2\ Hz)

Explanation:

From the question we are told that

  The  length of the ear canal is  l = 1.3 \ cm  =\frac{1.3}{100}  =  0.013 \ m

   The  speed of sound is assumed to be  v_s  =  344 \ m/s

Now  taking look at a typical  ear canal  we see that we assume it is  a  closed pipe

   Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

            f = \frac{v_s}{4 * l }

 substituting values  

          f = \frac{344}{4 * 0.013 }

         f = 6615.4 \ Hz

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

        f_1 =  \frac{3v_s}{4 * l}

 substituting values  

       f_1 =  \frac{3 *  344}{4 * 0.013}

       f_1 =   19846.2 \ Hz

Given that sound would be loudest in the pipe at the frequency, it implies that the child  will have an increased audible sensitivity at this  frequencies

6 0
3 years ago
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