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3 years ago

12

A small bar of pure gold whose density is 19.3g/cm. Displaces 80 cm

1 answer:

nasty-shy [4]3 years ago

3 0

Answer:

The mass of the gold bar is 1,544 g

Explanation:

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2 years ago

Suppose that a comet that was seen in 550 A.D. by Chinese astronomers was spotted again in year 1941. Assume the time between ob

pickupchik [31]

To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.

Kepler's third law tells us that

$T^2 = \frac{4\pi^2}{GM}a^3$

Where

T= Period

G= Gravitational constant

M = Mass of the sun

a= The semimajor axis of the comet's orbit

The period in years would be given by

$T= 1941-550\\T= 1391y(\frac{31536000s}{1y})\\T=4.3866*10^{10}s$

PART A) Replacing the values to find a, we have

$a^3= \frac{T^2 GM}{4\pi^2}$

$a^3 = \frac{(4.3866*10^{10})^2(6.67*10^{-11})(1.989*10^{30})}{4\pi^2}$

$a^3 = 6.46632*10^{39}$

$a = 1.86303*10^{13}m$

Therefore the semimajor axis is $1.86303*10^{13}m$

PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by

$R = a(1-e)$

$R = 1.86303*10^{13}(1-0.997)$

$R= 5.58*10^{10}m$

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3 years ago

Consider a 40,000 km steel pipe in the shape of a ring that fits snuggly all around the circumference of the Earth. We are heati

Tatiana [17]

Answer:

The Height is H = 70.02 m

Explanation:

We are given that the

Initial length is = $40000\ Km$ = $40,000 *10^{3} m$

from what we are told in the question the circumference of the circle is = $40,000 Km$

This means that the Radius would be :

Let C denote the circumference

So

$C = 2 \pi r$

=> $r = \frac{C}{2 \pi}$

$r = \frac{40,000}{2 \pi } = \frac{40,000*10^{3}}{2 *3.142} = 6.365*10^6 m$

We are told that 1-meter bar of steel that increases its temperature by 1 degree C will expand $11*10^{-6}$ meters

Hence

The final length would be

$40000*10^3 *(T + \alpha )$

Where T is the change in temperature $\alpha$ is the Coefficient of linear expansion for steel

let $L_{final}$ denote the final length

So

$L_{final} =40000*10^{6} *[1+ 11*10^{-6}]$

$= 40000440 \ m$

Now the Height is mathematically represented as

$Height(H) \ = \frac{change \ in \ radius \ }{2 \pi}$

$= \frac{(40000440-40000*10^3)}{2*3.142}$

$= 70.02m$

3 0

3 years ago