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BartSMP [9]
3 years ago
12

Just for fun, a person jumps from rest from the top of a tall cliff overlooking a lake. in falling through a distance h, she acq

uires a certain speed v. assuming free-fall conditions, how much farther must she fall in order to acquire a speed of 2v? express your answer in terms of h.
b. would the answer to part
a. be different if this event were to occur on another planet where the acceleration due to gravity had a value other than 9.80 m/s2? explain.
Physics
1 answer:
RideAnS [48]3 years ago
4 0
For this we need to use several formulas:
v = a*t
This simply means that after 1 second we will have some speed and if we double time, speed will double as well.

But, relation time-traveled distance isn't linear. Traveled distance we calculate:
s = 1/2*a*t^2                we can say that h=s
so if we double time, traveled distance will increase 4 times because of square relation between time and traveled distance.

New traveled distance (h2) will be: h2=4*h

b)  Because distance depends on gravity acceleration in free fall h2 will certainly change but the relation h2=4*h will remain the same.
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An MRI scanner is based on a solenoid magnet that produces a large magnetic field. The magnetic field doesn't stop at the soleno
GaryK [48]

Answer:

The maximum change in  flux is \Delta \o = 0.1404 \ Wb

The average  induced emf     \epsilon =0.11232 V

Explanation:

   From the question we are told that

             The speed of the technician is v = 0.80 m/s

              The distance from the scanner is d = 1.0m

              The  initial magnetic field is  B_i = 0T

               The final magnetic field is B_f = 6.0T

                 The diameter of the loop is  D = 19cm = \frac{19}{100} = 0.19 m

The area of the loop is mathematically represented as

        A  =  \pi [\frac{D}{2} ]^2

             = 3.142 \frac{0.19}{2}

             = 0.02834 m^2

At maximum the change in magnetic field is mathematically represented as

            \Delta \o = (B_f - B_i)A

  =>      \Delta  \o = (6 -0)(0.02834)

                  \Delta \o = 0.1404 \ Wb

The  average induced emf is mathematically represented as

           \epsilon =  \Delta \o v

              = 0.1404 * 0.80

             \epsilon =0.11232 V

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