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denis-greek [22]
3 years ago
9

A object of mass 33 is dropped from a height of 85 meters. Calculate the average power developed by the object in falling throug

h this distance.
Take g as 9.81 m/s^2

give the answer in watts to three significant figures
Physics
1 answer:
DochEvi [55]3 years ago
3 0
Power  =  Work done  /  Time taken.

Work done =  mgh
Mass, m = 33kg    ( Am presuming it is 33 kg).
h =  85 m.
Work done =  33 * 9.81* 85 =  27517.05  J.

Time taken.
Since object was dropped from height, it fell under gravity.
Using    H =  ut  +  (1/2) * gt^2.              u = 0.
               H =  1/2 gt^2.
               t  =   (2H/g) ^ (1/2)
               t  =  (2*85/9.81) ^ 0.5 =  4.1628 s.

Power  =    27517.05 / 4.1628 = 6610.23 Watts.
             =  6610 W  to  3 S. f.

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The units of G must be C. m³ / ( kg s² )

<h3>Further explanation</h3>

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }

<em>F = Gravitational Force ( Newton )</em>

<em>G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )</em>

<em>m = Object's Mass ( kg )</em>

<em>R = Distance Between Objects ( m )</em>

Let us now tackle the problem !

To find unit of Gravitational Constant can be carried out in the following way:

F = G \frac{m_1 ~ m_2}{R^2}

{[N]}= G\frac{{[kg]}{[kg]}}{{[m^2]}}

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G = \frac{{[kg ~ m / s^2]}{[m^2]}} {{[kg^2]} }

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G = \frac{{[m^3 / s^2]}} {{[kg]} }

\boxed {G = \frac{{[m^3]}} {{[kg ~ s^2]} }}

The unit of G must be \large {\boxed {\frac{m^3} {kg ~ s^2 }}}

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

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