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IRISSAK [1]
3 years ago
12

Pluto was first observed in 1930, and its largest moon, Charon, was discovered in 1978. A few years after Charon’s discovery, as

tronomers were able to observe a series of eclipses as Pluto and Charon passed in front of one another. By studying how the brightness of Pluto and Charon changed as they eclipsed each other, astronomers were able to measure the masses and radii of both Pluto and its moon. What did these measurements imply about the average densities of Pluto and Charon?
Physics
1 answer:
mylen [45]3 years ago
8 0

Answer:

The average densities of both matches the expected density for objects made from water ice.

Explanation:

Charon's density is 1.2 to 1.3 g / cm3, while Pluto's density is 1.8 to 2.1 g / cm3. This was discovered in many researches and measurements of these two celestial bodies, with the objective of understanding them and promoting efficient scientific knowledge.

With the measurements of the average densities between pluto and Charon it was possible to conclude several statements about them. Firstly, it is possible to see that the two formed independently and at different times, in addition to indicating the existence of few rocks in charon, which is consistent with the average density of objects made mostly of water ice.

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A meter stick is free to rotate about an axis through one of its end. Find the force F needed to balance this meter stick if the
Mumz [18]

Answer:

Explanation:

Component of force perpendicular to stick

= F Sin 60°

=√3 / 2 F.

Taking torque about the other end

= √3 / 2 F x 1 Nm

Weight of stick = 60 gm

= 60 x 10⁻³ kg

= 60 x 10⁻³ x 9.8 N

= .588 N

This weight will act from the middle point of stick so torque about the

other end

= .588 x 1 Nm

Balancing these two torques we have

.588 = √3 /2 F

F=\frac{2\times0.588}{\sqrt{3} }

F = 0.679 N

6 0
3 years ago
Describe what it means if a chemical or physical property is periodic
vesna_86 [32]
It means that you consider the elements as a list organized by atomic number, the property is seen to repeat over and over as you move through that list.
6 0
3 years ago
A man-made satellite orbits the earth in a circular orbit that has a radius of 32000 km. The mass of the Earth is 5.97e 24 kg. W
Gwar [14]

Answer:

= 3521m/s

The tangential speed is approximately 3500 m/s.

Explanation:

F = m * v² ÷ r

Fg = (G * M * m) ÷ r²

(m v²) / r = (G * M * m) / r²

v² = (G * M) / r

v = √( G * M ÷ r)

G * M = 6.67 * 10⁻¹¹ * 5.97 * 10²⁴ = 3.98199 * 10¹⁴

r = 32000km = 32 * 10⁶ meters  

G * M / r = 3.98199 * 10¹⁴ ÷ 32 * 10⁶

v = √1.24 * 10⁷  

v = 3521.36m/s

The tangential speed is approximately 3500 m/s.

8 0
3 years ago
Read 2 more answers
Helppppppppppppppppppppppppp
andrew11 [14]

Answer:

value of x=35°

y=72.5°

hope it helps you

make me brainliest plz

6 0
3 years ago
Problem 1: Three beads are placed along a thin rod. The first bead, of mass m1 = 24 g, is placed a distance d1 = 1.1 cm from the
Svet_ta [14]

Answer:

b)  x_{cm} = 4.88 cm , c) x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃) and d)

x_{cm}’= 1.88 cm

Explanation:

The definition of mass center is

    x_{cm} = 1/M ∑ xi mi

Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.

Part b

Apply this equation to our case.

Body 1

They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left

Body 2

They give us the mass (m₂ = 12.g) and the distance relative to the distance of the body 1, let's look for the distance from the left end (origin)

    D₂ = d₁ + d₂

    D₂ = 1.1 + 1.9

    D₂ = 3.0 cm

Body 3

Give the mass (m₃ = 56 g) and the position relative to body 2, let's find the distance relative to the origin

    D₃ = D₂ + d₂

    D₃ = 3.0 + 3.9

    D₃ = 6.9 cm

With this data we substitute and calculate the center of mass

    M = m₁ + m₂ + m₃

    M = 24 + 12 + 56

    M = 92 g

    x_{cm} = 1/92 (1.1 24 + 3.0 12 + 6.9 56)

    x_{cm} = 1/92 (448.8)

    x_{cm} = 4,878 cm

    x_{cm} = 4.88 cm

This distance is from the left end of the bar

Par c)

In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.

Body 1

It is at   d1 = -1.9 cm

It is negative for being on the left and the value is the relative distance of 1 to 2

Body 2

d2 = 0 cm

The reference system for her

Body 3

d3 = 3.9 cm

Positive because that is to the left of the reference system and is the relative distance between 2 and 3

Let's write the new center of mass (xcm')

    x_{cm} ’= 1/M  (m₁ d₁ + m₂ d₂ + m₃ d₃)

   

   x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃)

Part d) Let's calculate the value of the center of mass

    x_{cm}’= 1/92 ((24 (-1.9) +56 3.9)

    x_{cm}’= 1/92 (172.8)

    x_{cm}’= 1.88 cm

This distance is to the right of the central marble

3 0
3 years ago
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