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ankoles [38]
3 years ago
15

A large asteroid of mass 98700 kg is at rest far away from any planets or stars. A much smaller asteroid, of mass 780 kg, is in

a circular orbit about the first at a distance of 201 meters as a result of their mutual gravitational attraction. What is the speed of the second asteroid?
Physics
1 answer:
Citrus2011 [14]3 years ago
7 0

Answer:

1.81 x 10^-4 m/s

Explanation:

M = 98700 kg

m = 780 kg

d = 201 m

Let the speed of second asteroid is v.

The gravitational force between the two asteroids is balanced by the centripetal force on the second asteroid.

\frac{GMm}{d^{2}}=\frac{mv^2}{d}

v=\sqrt{\frac{GM}{d}}

Where, G be the universal gravitational constant.

G = 6.67 x 10^-11 Nm^2/kg^2

v=\sqrt{\frac{6.67 \times 10^{-11}\times 98700}{201}}

v = 1.81 x 10^-4 m/s

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Answer:

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2 years ago
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Les propriétés de l’air?
crimeas [40]

Answer:

Lorsque l'on détend l'air son volume augmente et sa pression diminue. L'air qui est un mélange de gaz est compressible et expansible. – Lorsque l'on comprime l'air, son volume diminue et sa pression augmente. – Lorsque l'on détend l'air, son volume augmente et sa pression diminue.

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2 years ago
On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-f
aliya0001 [1]

Answer:

15.3 s and 332 m

Explanation:

With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon

    gm = 1/6 ge

    gm = 1/6  9.8 m/s² = 1.63 m/s²

We calculate the range

    R = Vo² sin 2θ  / g

    R = 25² sin (2 30) / 1.63

    R= 332 m

We will calculate the time of flight,

   Y = Voy t – ½ g t2  

   Voy = Vo sin θ

When the ball reaches the end point has the same initial  height Y=0

0 = Vo sin  t – ½  g t2

0 = 25 sin (30)  t – ½ 1.63 t2

0= 12.5 t –  0.815 t2

We solve the equation

0= t ( 12.5 -0.815 t)

 t=0 s

t= 15.3 s

The value of zero corresponds to the departure point and the flight time is 15.3 s

Let's calculate the reach on earth

R2 = 25² sin (2 30) / 9.8

R2 = 55.2 m

R/R2 = 332/55.2

R/R2 = 6

Therefore the ball travels a distance six times greater on the moon than on Earth

5 0
3 years ago
Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final
Serhud [2]

Complete question:

Consider the hypothetical reaction 4A + 2B → C + 3D

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?

Answer:

the final concentration of A is 0.992 M.

Explanation:

Given;

time of reaction, t = 4.0 s

rate of change of the concentration of B =  -0.0760 M/s

initial concentration of A = 1.600 M

⇒Determine the rate of change of the concentration of A.

From the given reaction: 4A + 2B → C + 3D

2 moles of B ---------------> 4 moles of A

-0.0760 M/s of B -----------> x

x = \frac{4(-0.076)}{2} \\\\x = -0.152 \ M/s

⇒Determine the change in concentration of A after 4s;

ΔA = -0.152 M/s  x 4s

ΔA = -0.608 M

⇒ Determine the final concentration of A  after 4s

A = A₀ + ΔA

A = 1.6 M + (-0.608 M)

A = 1.6 M - 0.608 M

A = 0.992 M

Therefore, the final concentration of A is 0.992 M.

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