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Elis [28]
2 years ago
8

Which is the last stage in the life cycle of an average star? (1 point)

Physics
1 answer:
Shkiper50 [21]2 years ago
4 0

Answer:

The answer is A. White Dwarf

Explanation:

  • <u>average stars like sun end their life as white dwarf surrounded by disappearing planetary nebula. </u>
You might be interested in
What type of material is wrapped around an object could cause the object to lose the most heat
Dmitrij [34]
<span>A sheet of copper could cause the object to lose the most amount of heat. Copper is an essential element and a good conductor of heat. Heat can transfer from one end of a piece of copper to the other end.</span>
7 0
3 years ago
) Water falls from a height of 60m at the rate of 15kg/s to operate a turbine. The losses due to frictional force are 10% of ene
Angelina_Jolie [31]

Answer:

8100W

Explanation:

Let g = 10m/s2

As water is falling from 60m high, its potential energy from 60m high would convert to power. So the rate of change in potential energy is

P = \dot{E} = \dot{m}gh = 15*10*60 = 9000 J/s or 9000W

Since 10% of this is lost to friction, we take the remaining 90 %

P = 9000*90% = 8100 W

3 0
3 years ago
A ball rolls with a speed of 2.0m/s across a level table that is 1.0m above the floor. Upon reaching the edge of the table it fo
Mekhanik [1.2K]

Answer:0.58 m

Explanation:

The initial velocity of the ball is u = 2.0 m/s

The height of the table is, h = 1.0 m

The ball falls in vertical direction under acceleration due to gravity.

Time taken for ball to hit the floor:

h= ut + 0.5gt² ( from the equation of motion)

1.0 m=2.0 m/s × t+0.5 × 9.8 m/s²× t²

Solving this for t,

t = 0.29 s ( we have neglected the negative value of t)

In the same time, the ball would cover a horizontal distance of :

s = u t

⇒s = 2.0 m/s×0.29 s = 0.58 m

Thus, the landing spot is 0.58 m away from the table.

6 0
3 years ago
Read 2 more answers
A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. a
USPshnik [31]
Refer to the diagram shown below.

Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force

The applied torque is
T = F*r
   = (260 N)*(0.33 m)
   = 85.8 N-m

By definition,
T = I*α

Therefore,
I = T/α
  = (85.8 N-m)/(0.81 rad/s²)
  = 105.93 kg-m²

Answer: 105.93 kg-m²

6 0
3 years ago
--
enot [183]

Answer: 3 square feet

Explanation: I took the test

6 0
3 years ago
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