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3 years ago

9. Suppose you want to calculate how much

Physics

1 answer:

nikklg [1K]3 years ago

7 0

Answer:

The correct option is;

b. How high the barbell is being lifted

Explanation:

The work done in lifting the 160 N barbell is equal to the potential energy gained by the barbell in the final elevated position of the barbell

Therefore, in order to find the work done to lift the barbell, we can calculate the potential energy gained by the barbell when it is placed at height

The work done to lift the barbell = The potential energy, P.E., gained by the barbell

The potential energy of an object, P. E. = The mass of the object, m × The acceleration due to gravity, g × The height to which the barbell is lifted, h

∴ Mathematically, P.E. = m × g × h

Weight, W = Mass, m × The acceleration due to gravity, g

The weight, W of the barbell is given as W = 160 N

∴ W = m × g = 60 N

From which we have;

P.E. = m × g × h = W × h

∴ The work done to lift the barbell = P.E. = 60 N × h = 60 N × The height to which the barbell is lifted.

Therefore the information which need be known is the height to which the barbell.

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son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

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R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

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Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

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$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

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$2r = 2*10^3m$

Therefore

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$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

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4 years ago

ANSWER THIS QUESTION!!!! PLSSSSS!!! NEED HELP!!!!! 10 POINTS

Olegator [25]

Answer:

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Elena L [17]

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2 years ago

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