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Likurg_2 [28]
3 years ago
11

A comet is in an elliptical orbit around the sun. its closest approach to the sun is a distance of 4.5 1010 m (inside the orbit

of mercury), at which point its speed is 9 104 m/s. its farthest distance from the sun is far beyond the orbit of pluto. what is its speed when it is 6 1012 m from the sun?
Physics
1 answer:
barxatty [35]3 years ago
4 0

r1 = 5*10^10 m , r2 = 6*10^12 m

v1 = 9*10^4 m/s

From conservation of energy

K1 +U1 = K2 +U2

0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2

0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

v2 = 5.35*10^4 m/s

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Answer:

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(b)

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2 2 2.00 0.178 mm  200 g , T  0.250 s,E 2.00 J

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(b)

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Explanation:

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8 0
3 years ago
Un pintor de 75.0 kg sube por una escalera de 2.75 m que está inclinada contra una pared vertical. La escalera forma un ángulo d
dezoksy [38]

Answer:

Work done, W = 1786.17J

Explanation:

The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "

Mass of a painter, m = 75 kg

He climbs 2.75-m ladder that is leaning against a vertical wall.

The ladder makes an angle of 30 degrees with the wall.

We need to find the work done by the gravity on the painter.

The angle between the weight of the painter and the displacement is :

θ = 180 - 30

= 150°

The work done by the gravity is given by :

W=Fd\cos\theta\\\\=75\times 10\times 2.75\times \cos30\\\\=1786.17\ J

Hence, the required work done is 1786.17 J.

6 0
2 years ago
A student tries to produce notes of higher frequency by blowing a trumpet harder. he will succeed?​
sdas [7]

Answer:

Please mark as Brainliest!!

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He will not succeed as it blows the trumpet harder that only increase the amplitude of the sound wave that means the intensity of the sound will increase but the frequency remains the same.

7 0
2 years ago
Read 2 more answers
A 500 kg table with 4 legs rests on solid flat ground. I place 3 books on the center of the table with masses of 10 kg, 20 kg an
BARSIC [14]

Answer:

1,373.4 N

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The mass of the table acts at the centre in addition to the books since that is the centre of gravity of the table.

Mass of books will be 10kg+20kg+30kg=60 kg

Total mass of table and books will be 500kg+60kg=560 kg

This mass is evenly distributed into the four legs hence 560kg/4 legs=140 kg per leg

Force is product of mass and acceleration due to gravity hence F=gm

Taking g as 9.81 m/s2 then

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6 0
3 years ago
A ball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the i
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The position of the first ball is

y_1=h-\dfrac g2t^2

while the position of the second ball, thrown with initial velocity v, is

y_2=vt-\dfrac g2t^2

The time it takes for the first ball to reach the halfway point satisfies

\dfrac h2=h-\dfrac g2t^2

\implies\dfrac h2=\dfrac g2t^2

\implies t=\sqrt{\dfrac hg}

We want the second ball to reach the same height at the same time, so that

\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2

\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)

\implies h=v\sqrt{\dfrac hg}

\implies v=\sqrt{hg}

8 0
3 years ago
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