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Flura [38]
3 years ago
10

Listed following are several astronomical objects. Rank these objects based on their diameter, from largest to smallest. (Note t

hat the neutron star and black hole in this example have the same mass to make your comparison easier, but we generally expect black holes to have greater masses than neutron stars.)
a.main-sequence star of spectral type a.
b. jupiter.
c. a one solar mass white dwarf.
d. the moon.
e. a two solar mass neutron star.
f. event horizon of a two solar mass black hole.
Physics
1 answer:
daser333 [38]3 years ago
3 0

Answer:

Ranked

Explanation:

The rank of these objects from largest to smallest based on diameter is as follows

1. main-sequence star of a spectral type A

2. Jupiter

3. a one-solar-mass white dwarf

4.the Moon

5. a two-solar-mass neutron star

6. the event horizon of a two-solar-mass black hole

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Please help!!
sukhopar [10]

Answer:

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<u>ANSWER</u><u>;</u>

The force exerted by the thrusters caused the pod to change direction.

WHAT NEW THEORIES DO YOU HAVE?

ANSWER;

This pod moved differently because it was more massive.

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3 0
2 years ago
A frictionless spring with a 3-kg mass can be held stretched 0.8 meters beyond its natural length by a force of 40 newtons. If t
Free_Kalibri [48]

Answer:

Explanation:

mass m = 3 kg

spring constant be k

k x .8 = 40 N

k = 40 / .8 = 50 N /m

angular frequency ω = √ ( k / m )

= √ ( 50 / 3 )

= 4.08 rad /s

Let amplitude of oscillation be A .

1/2 k A² = 1/2 m v²

50 A² = 3 x 1²

A = .245 m = 24.5 cm

For displacement , the equation of SHM is

x = A sinωt

= 24.5 sin4.08 t

x = 24.5 sin4.08 t

Here, angle 4.08 t is in radians .

3 0
3 years ago
(10p+15,15-10q)=(25,5)​
Harrizon [31]
63783626736377474737377447
8 0
2 years ago
How does friction affect the distance of an object?
stiv31 [10]
It slows the object down so it cannot move well and evetually the object cannot be pushed and farther
4 0
3 years ago
A bird uses 10 N of force to pull a worm out of the ground a distance of 3 inches. How much work did the bird do?
pochemuha

Answer:

The work done by the bird is 0.762 J

Explanation:

Given;

force applied by the bird, f = 10 N

distance the bird moved the worm, d = 3 inches = 0.0762 m

The work done by the bird is given by;

W = F x d

where;

W is the work done by the bird

d is the distance the bird moved the load

Substitute the given values and estimate the work done by the bird;

W = 10 x 0.0762

W = 0.762 J

Therefore, the work done by the bird is 0.762 J

6 0
3 years ago
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