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egoroff_w [7]
3 years ago
8

What happens to a star after it goes supernova? Does the remnant of the star form a dwarf star, neutron star, black hole, or som

ething else? Thanks!!!
Physics
2 answers:
qaws [65]3 years ago
8 0
I believe it forms a black hole. Eek!
il63 [147K]3 years ago
8 0
Well after the star goes supernova, It becomes a Neutron Star, an extrememly powerful and dense Explosion full of atoms and neutrons, if the mass does not exceed than it will remain to be a neutron star. Hope that helps
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juin [17]

Answer:

b

Explanation:

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Arisa [49]
The last one, handmade gifts require more of the givers time!
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A ring, cylinder, solid sphere, and hollow sphere are all released from rest from the same height on an inclined surface, at the
crimeas [40]

Answer:

Explanation:

  • The expression for acceleration of the rolling body on an inclined plane is given as a = gsinФ/1 + k²/R²
  • where Ф is the angle of inclination, R is the radius, k is the radius of gyration.
  • The potential energy of the system is given as ; PE = mgh
  • The potential energy will be constant for ring, cylinder, solid sphere, and hollow sphere.
  • The total kinetic energy of the rolling body is ; KE = mv²/2 + Iw²/2
  • Hence, the total kinetic energy of the ring, cylinder, solid sphere and hollow sphere will be constant.

2. The moment of inertia of the ring is given as ;

I = mR²

The moment of inertia of the ring is maximum and therefore reaches the bottom last.

7 0
3 years ago
A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a
il63 [147K]

Answer:

0.0000109261200583 s

0.0109261200583

Explanation:

d_2 = Distance from right ear = 3 m

s = Distance between ears = 15 cm

v = Speed of sound in air = 343 m/s

Distance between the left ear and the bird

d_1=\sqrt{s^2+d_2^2}\\\Rightarrow d_1=\sqrt{0.15^2+3^2}\\\Rightarrow d_1=3.00374765918\ m=3.004\ m

Time

t=\dfrac{Distance}{Speed}

Time difference would be

\Delta T=\dfrac{d_1}{v}-\dfrac{d_2}{v}\\\Rightarrow \Delta T=\dfrac{3.00374765918}{343}-\dfrac{3}{343}\\\Rightarrow \Delta T=0.0000109261200583\ s

The time difference is 0.0000109261200583 s

Time period is given by

T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{1000}\\\Rightarrow T=10^{-3}\ s

The ratio is

\dfrac{\Delta T}{T}=\dfrac{0.0000109261200583}{10^{-3}}\\\Rightarrow \dfrac{\Delta T}{T}=0.0109261200583

The ratio is 0.0109261200583

7 0
3 years ago
A 1.60 m tall person lifts a 2.10-kg book from the ground so it is 2.20 m above the ground. What is the potential energy of the
Viefleur [7K]

Answer:

Explanation:

Given

mass of book(m)=2.1 kg

height up to which book is lifted is (h)2.2 m

height of person (h_0)1.6 m

Potential energy of book relative to ground=mgh

PE=2.1\times 9.8\times 2.2=45.276 J

(b)PE w.r.t to person head =mg(h-h0)

=2.1\times 9.8\times (2.2-1.6)=12.348 J

work done by person in lifting box 2.2 m w.r.t floor

Word done =Potential Energy of box relative to floor=45.2 J

6 0
3 years ago
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