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3 years ago

If the sun is scaled down to 9.5 inches in diameter, the earth would be about _____ inches in diameter

Physics

1 answer:

vovangra [49]3 years ago

7 0

Answer: 0.087 inches

Explanation:

This problem can be solved applying the Rule of three, which is a mathematical rule to find out an amount that is with another quantity given in the same relation as other two also known.

In this sense, the actual diameter of the Sun is 1,391,020 km and the actual diameter of the Earth is 12,742 km.

Now, applying the Rule of three:

Sun 1,391,020 km ------ 9.5 inches

Earth 12,742 km --------- ? inches

$? inches=\frac{(12,742 km)(9.5 inches)}{1,391,020 km}$

$? inches=0.087 inches$

Therefore:

If the sun is scaled down to 9.5 inches in diameter, the earth would be about 0.087 inches in diameter.

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3 years ago

Assume that Parker Company will receive SF200,000 in 360 days. Assume the following interest rates:

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Answer:

b. $96,914

Explanation:

360-day borrowing rate = 5%

spot rate = 0.48

360-day deposit rate = 6%

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3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

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

G =Gravitational Universal Constant

R = Distance /Radius

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

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

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

$dKE = -dU$

Therefore replacing we have that,

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

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

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

$v = 816.7m/s$

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

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

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

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

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

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

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3 years ago

a pool ball leaves a 0.60-meter high table with an initial high table with an initial horizontal velocity of 2.4m/s. what is the

inysia [295]

Answer:

0.84 m

Explanation:

Given in the y direction:

Δy = 0.60 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

0.60 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.35 s

Given in the x direction:

v₀ = 2.4 m/s

a = 0 m/s²

t = 0.35 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (2.4 m/s) (0.35 s) + ½ (0 m/s²) (0.35 s)²

Δx = 0.84 m

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4 years ago