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nata0808 [166]
2 years ago
15

Waldo gets stopped by the police, and the police officer tells Waldo that she caught him on radar doing 50 miles per hour in a 4

0 miles per hour zone.
Waldo responds, " I was traveling 50 miles per hour. I sat at home for 1 hour and then drove for 1 hour before you stopped me. I traveled 50 miles total in that second hour. So, my speed was 25 miles per hour. That's well under the speed limit."
Waldo is wrong, per speeding laws, and does receive a speeding ticket. Explain how Waldo came up with a speed of 25 miles per hour in multiple sentences.
Physics
2 answers:
sergij07 [2.7K]2 years ago
5 0

Answer:

He calculated his average speed.

Explanation:

Waldo calculated an average of his speed in 2 hours rather than his speed whilst driving which means instead of doing 50/1 hour = 50 mph, he did 50/2hours which is how he got 25 mph. Hope this helps!

slava [35]2 years ago
5 0
50/2 hours=25
I think this is correct!
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A crane used 250,000 Joules of work to move a beam to the top of a building in 20 seconds. How much power did the crane use?
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Answer:

12500W

Explanation:

Given parameters:

Work done  = 250000J

Time taken  = 20s

Unknown:

Power of the crane = ?

Solution:

Power is the defined as the rate at which work is being done;

  Mathematically;

        Power = \frac{work done}{time }

 insert the parameters and solve;

      Power  = \frac{250000}{20}   = 12500W

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Explanation:

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Sphere A with mass 80 kg is located at the origin of an xy coordinate system; sphere B with mass 60 kg is located at coordinates
IRINA_888 [86]

Answer:

Fc = [ - 4.45 * 10^-8 j ] N  

Explanation:

Given:-

- The masses and the position coordinates from ( 0 , 0 ) are:

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       Sphere B : ma = 60 kg , ( 0.25 , 0 )

       Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15

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what is the gravitational force on C due to A and B?

Solution:-

- The gravitational force between spheres is given by:

                       F = G*m1*m2 / r^2

Where, r : The distance between two bodies (sphere).

- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-

 Determine the angle (α) between vectors rac and rab using cosine rule:

                   cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}

 Determine the angle (β) between vectors rbc and rab using cosine rule:

                   cos ( \beta  ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta  ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta  ) = 0.6\\\\\beta  = 53.13^{\circ \:}

- Now determine the scalar gravitational forces due to sphere A and B on C:

       Between sphere A and C:

                  Fac = G*ma*mc / rac^2

                  Fac = (6.674×10−11)*80*0.2 / 0.2^2  

                  Fac = 2.67*10^-8 N

                  vector Fac = Fac* [ - cos (α) i + - sin (α) j ]

                  vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]

                  vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N

       Between sphere B and C:

                  Fbc = G*mb*mc / rbc^2

                  Fbc = (6.674×10−11)*60*0.2 / 0.15^2  

                  Fbc = 3.56*10^-8 N

                  vector Fbc = Fbc* [ cos (β) i - sin (β) j ]

                  vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]

                  vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N

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                  Fc = [ - 2.136 i - 1.602 j ]*10^-8  + [ 2.136 i - 2.848 j ]*10^-8

                  Fc = [ - 4.45 * 10^-8 j ] N  

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