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4 years ago

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

1 answer:

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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Soap bubble interference Light of 690-nm wavelength interferes constructively when reflected from a soap bubble having refractiv

Nikitich [7]

Answer:

The two possible thicknesses of the soap bubble is 129 nm and 389 nm.

Explanation:

Given that,

Wavelength = 690 nm

Refractive index = 1.33

We need to calculate the two possible thicknesses of the soap bubble

Using formula of thickness

$t=\dfrac{(m+\dfrac{1}{2})\lambda}{2n}$

For m = 0,

Put the value into the formula

$t=\dfrac{\dfrac{690\times10^{-9}}{2}}{2\times1.33}$

$t=129\times10^{-9}\ m$

$t=129\ nm$

For m=1,

Put the value into the formula

$t=\dfrac{\dfrac{3\times690\times10^{-9}}{2}}{2\times1.33}$

$t=389\times10^{-9}\ m$

$t=389\ nm$

Hence, The two possible thicknesses of the soap bubble is 129 nm and 389 nm.

5 0

3 years ago

A 15-kN tensile load will be applied to a 50-m length of steel wire with E = 200 GPa. Determine the smallest diameter wire that

sasho [114]

Answer:

d=13.81 mm

Explanation:

Given that

P = 15 KN ,L = 50 m

E= 200 GPa

ΔL = 25 mm

σ = 150 MPa

Lets take d=Diameter

There are we have two criteria to find out the diameter of the wire

Case I :

According to Stress ,σ = 150 MPa

P = σ A

$A=\dfrac{P}{\sigma}$

$d=\sqrt{\dfrac{4P}{\pi \sigma}}$

By putting the values

$d=\sqrt{\dfrac{4\times 15000}{\pi \times 150}}$

d= 11.28 mm

Case II:

According to elongation ,ΔL = 25 mm

$\Delta L=\dfrac{PL}{AE}$

$A=\dfrac{PL}{E\Delta L}$

$A=\dfrac{4PL}{\pi E\Delta L}$

$d=\sqrt{\dfrac{4\times 15000\times 50000}{\pi \times 200\times 1000\times 25}}$

d=13.81 mm

Therefore the answer will be 13.81 mm .Because it satisfy both the conditions.

3 0

3 years ago

Parker (73.2 kg) is being dragged down the hall with an applied force of 123 N. If the frictional force is 27.4 N, what is the c

levacccp [35]

Answer:

The coefficient of friction in the hall is 0.038

Explanation:

Given;

mass of the Parker, m = 73.2 kg

applied force on the parker, F = 123 N

frictional force, Fs = 27.4 N

the coefficient of friction in the hall = ?

frictional force is given by;

Fs = μN

Where;

μ is the coefficient of friction

N is normal reaction = mg

Fs = μmg

μ = Fs / mg

μ = (27.4) / (73.2 x 9.8)

μ = 0.038

Therefore, the coefficient of friction in the hall is 0.038

6 0

3 years ago

How does regular exercise help older adult stay healthy?

gtnhenbr [62]

It’s helps the heart stay healthier by respiratory of the heart

5 0

3 years ago

Read 2 more answers

Why does it take mars over 500 days to orbit the sun?

kakasveta [241]

Answer:

Mar's orbital path is more than that of Earth, thus it takes more number of days to orbit around the sun.

Explanation:

Mars takes over 500 days to orbit all the way around the sun than Earth because its distance from the sun (228 million kilometers) is greater than that of Earth (150 million kilometers) which takes it 365 days.

Planets that orbit closer to the sun take shorter time to orbit around the sun because the cover a shorter orbital distance and orbit faster than those planets further from the sun.

<u>For example</u>

Using Earth's distance from the sun, 150 million kilometers and the number of days taken to orbits the sun ,365 days and the distance Mars is from the Earth, 228 million kilometers, you can approximate the time Mar takes to orbit the sun as:

Earth 150 million kilometers = 365 days

Mars 228 million kilometers= ?

Cross product ; (228 *365) /150 =555 -----(a value closer to that in the question)

6 0

3 years ago