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cestrela7 [59]
4 years ago
7

E4.Suppose Galileo’s pulse rate was 75 beats per minute. How many beats per second is this? What is the time in seconds between

consecutive pulse beats? How far (in meters) does an object fall in this time when dropped from rest? What is this distance in feet (use the conversion factors on the inside front cover)?
Physics
1 answer:
Umnica [9.8K]4 years ago
3 0

Answer

given,

Galileo's pulse rate = 75 beats per minutes

Beats per second

1 minutes = 60 s

pulse rate = \dfrac{75}{60}

                  = 1.25 beats/s

times in second

t = \dfrac{1}{1.25}

      t = 0.8 s/pulse

distance of the object

initial velocity = 0 m/s

s = ut + \dfrac{1}{2}gt^2

s = \dfrac{1}{2}\times 9.8\times 0.8^2

      s = 3.14 m

distance in feet

  1 m = 3.28 ft

 3.14 m = 3.14 x 3.28 ft

             = 10.3 ft

distance in feet is equal to 10.3 ft.

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The time that is required for a vibrating object to complete one full cycle is called : Group of answer choices
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Answer:

Correct option : (c) period.                        

Explanation:

The time that is required for a vibrating object to complete one full cycle is called the time period. If f is the frequency of a wave, then the relation between the frequency and the time period is given by :

T=\dfrac{1}{f}

These are the characteristics of a wave. Some other characteristics are wavelength, amplitude, intensity etc. So, the correct option is (c) "period".

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4 years ago
During a takeoff run, an aircraft starts from rest and has a lift-off speed of 120 km/h. a. What minimum constant acceleration d
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Answer:

1.984 m/s^2

Explanation:

initial velocity of air craft, u = 0 m/s

final speed of the aircraft, v = 120 km/h

Convert the speed into m/s from km/h

So, v = 120 km/h = 33.33 m/s

distance, s = 280 m  

Let a be the acceleration of the aircraft.

Use third equation of motion

v^{2}=u^{2}+2\times a\times s

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a = 1.984 m/s^2

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4 years ago
An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
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Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

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