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Korolek [52]
3 years ago
13

If a person is walking at 1.2 m/s and 60 seconds later the person is running at 10 m/s, what was the acceleration rate?​

Physics
1 answer:
Marina CMI [18]3 years ago
6 0
The acceleration rate would be .14667 m/s^2
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Amy throws a softball through the air. What are the different forces acting on the ball while it’s in the air?
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An applied force<span> is a </span>force<span> that is </span>applied<span> to an object by a person or another object.
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4 0
3 years ago
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A wave can be best defined as <br><br> P.S pls help
Sidana [21]

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I hope this helped you :)

4 0
3 years ago
The center of mass of the system is the point at which the total mass of the system could be concentrated without changing the _
Sveta_85 [38]

  The centre of mass of the system is the point at which the total

  mass of system could be concentrated without changing the moment

  of the system.

Centre of Mass is the point at which the whole mass of the system

is assumed to be concentrated.

 The general formula for the COM is:

              xₙ =  Σmₐxₐ / Σmₐ         where,  a = 1,2,3.........n

  Here the term Σ mₐ xₐ is called the first moment of the system and the

  denominator expression is called total mass of the system.

    Therefore, from this theory we can say that the moment of the system

    remain unchanged while calculating the COM.

  Hence, The centre of mass of the system is the point at which the total

  mass of system could be concentrated without changing the moment

  of the system.

   learn more about centre of mass here:

             brainly.com/question/3454419

                #SPJ4

8 0
2 years ago
A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm
iVinArrow [24]

Answer:

m = 14*26 = 364

Explanation:

overall magnification is given as m

m = m_{o}* m_{e}

mo magnification of objective lens

me magnification of EYE lens

where mo is given as

m_{o} = \frac{v_{o}}{-u _{0}}

and me as

m_{e} = 1+\frac{D}{f_{e}}

d is distant of distinct vision = 25.0 cm for normal eye

fe =  focal length of eye piece

focal length of objective lense is 0.140 cm

we know that

\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{u_{0}} + \frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{0.150} + \frac{1}{0.14}

\frac{1}{v_{o}} = 2.1 cm

m_{o} = \frac{2.1}{0.150} = 14

m_{e} = 1+\frac{25}{1}

m_{e} =26

m = m_{o}* m_{e}

m = 14*26 = 364

4 0
3 years ago
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