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Sergeu [11.5K]
3 years ago
7

If your chunk of gold weighed 1 N in which case would you have the largest mass of gold?

Physics
1 answer:
kotykmax [81]3 years ago
3 0
Ah ha !  Very interesting question.
Thought-provoking, even.

You have something that weighs 1 Newton, and you want to know 
the situation in which the object would have the greatest mass.

          Weight = (mass) x (local gravity)

          Mass  =  (weight) / (local gravity)

          Mass  =  (1 Newton) / (local gravity)

"Local gravity" is the denominator of the fraction, so the fraction
has its greatest value when 'local gravity' is smallest.  This is the
clue that gives it away.

If somebody offers you 1 chunk of gold that weighs 1 Newton,
you say to him:

   "Fine !  Great !  Golly gee, that's sure generous of you.  
But before you start weighing the chunk to give me, I want you
to take your gold and your scale to Pluto, and weigh my chunk
there.   And if you don't mind, be quick about it."

The local acceleration of gravity on Pluto is  0.62 m/s² ,
but on Earth, it's 9.81 m/s.

So if he weighs 1 Newton of gold for you on Pluto, its mass will be
1.613 kilograms, and it'll weigh 15.82 Newtons here on Earth. 

That's almost 3.6 pounds of gold, worth over $57,000 !


It would be even better if you could convince him to weigh it on
Halley's Comet, or on any asteroid.  Wherever he's willing to go
that has the smallest gravity.  That's the place where the largest
mass weighs 1 Newton.

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Find a first-degree polynomial function p1 whose value and slope agree with the value and slope of f at x = c. f(x) = cot(x), c
Neporo4naja [7]

The correct answer is y=-2x+(1/2)

y = f'(x)· x + c

Y = -2x + C

1 = -2x π/4 + C

=) C = I + π/2

y=-2x+(1/2)  is the first-degree polynomial.

First-degree polynomials are the simplest polynomials. Here, we'll talk about a few qualities and connect the terms polynomial, function, and equation. Write a polynomial equation in standard form before attempting to solve it. Factor it, then set each variable factor to zero after it has reached zero. The original equations' answers are the solutions to the derived equations. Factoring cannot always be used to solve polynomial equations. For instance, the polynomial 2x+5 has an exponent of 1. The most typical kinds of polynomials used in algebra and precalculus are zero polynomial functions.

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7 0
1 year ago
A pendulum of length L=36.1 cm and mass m=168 g is released from rest when the cord makes an angle of 65.4 degrees with the vert
pychu [463]

(a) -0.211 m

At the beginning the mass is displaced such that the length of the pendulum is L = 36.1 cm and the angle with the vertical is

\theta=65.4^{\circ}

The projection of the length of the pendulum along the vertical direction is

L_y = L cos \theta = (36.1 cm)(cos 65.4^{\circ})=15.0 cm

the full length of the pendulum when the mass is at the lowest position is

L = 36.1 cm

So the y-displacement of the mass is

\Delta y = 15.0 cm - 36.1 cm = -21.1 cm = -0.211 m

(b) 0.347 J

The work done by gravity is equal to the decrease in gravitational potential energy of the mass, which is equal to

\Delta U = mg \Delta y

where we have

m = 168 g = 0.168 kg is the mass of the pendulum

g = 9.8 m/s^2 is the acceleration due to gravity

\Delta y = 0.211 m is the vertical displacement of the pendulum

So, the work done by gravity is

W=(0.168 kg)(9.8 m/s^2)(0.211 m)=0.347 J

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(c) Zero

The work done by a force is:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

In this situation, the tension in the string always points in a radial direction (towards the pivot of the pendulum), while the displacement of the mass is tangential (it follows a circular trajectory): this means that the tension and the displacement are always perpendicular to each other, so in the formula

\theta=90^{\circ}, cos \theta = 0

and so the work done is zero.

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