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Hatshy [7]
3 years ago
12

The order of evolutionary stages of a star like the Sun would be Main Sequence, giant, planetary nebula, and finally:

Physics
1 answer:
professor190 [17]3 years ago
6 0

Answer:

Option E

Explanation:

Red Dwarfs are abundant throughout the universe. They are stable and considered to be candidates for exoplanets capable of sustaining life.

White Dwarfs are stars in the last phase of their evolution. A red giant of under 10 solar masses that has used its critical elements will shed its outer layers, stats to collapse under its own gravity and shrink to a white dwarf. They longer experience nuclear fusion.

The two bodies we are most interested in are Giants and Supergiants but it is important to look at others.

Neutron Stars are collapsed giants that had an original mass many times that of the Sun and explode as a supernova leaving a neutron star. These stars are compressed so much that they are composed entirely of neutrons, parts of the atom without electrical charge. This is the equivalent of the size of the Sun in the same area as a city or the human population on Earth fitting inside an area the size of a sugar cube.

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an experiment is set up to measure the effect of a gasoline additive on fuel consumption rate. what would be the control in this
Tamiku [17]
The control setup in this experiment would be one tank that does not contain any of the additives. Since the tanks with the gasoline additives would need to be compared with a tank that is not affected by the results of these additives.
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3 years ago
Write a hypothesis about how the height of the cylinder affects the temperature of the water. Use the "if . . . then . . . becau
AnnyKZ [126]
The statement that can be used to answer this  question is:

"If the cylinder is brought higher then, its temperature when brought down becomes higher because a greater amount of potential energy is converted to thermal energy."

The potential energy is converted to thermal energy when the object is released the velocity becomes higher because of the acceleration due to gravity.
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4 years ago
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A woman pushes an oak chest across an oak floor at a constant speed of 0.450 m/s. The chest has a mass of 40.0 kg, and the coeff
natima [27]

Answer:

A. 243 N

Explanation:

Friction is the force that opposes the relative motion between systems that are in contact.

This friction force that opposes the motion of the oak chest across the oak surface will be equal and opposite to that exerted by the woman.

First find the normal force which is the force that would point directly upwards to support weight of the block.

Normal force, N= mg where m is the mass of the chest and g is the acceleration due to gravity.

Given m=40 kg and g=9.80 m/s²

N force=40×9.80 =392N

Then find the force of friction which is given by the formula;

<em>F=μN where μ is friction coefficient for the  oak chest  and N is the normal force on the chest</em>

Given <em>μ</em>=0.620  and N force = 392 N then it will be;

F=0.620× 392 =243.04 N

Answer : 243 N

6 0
3 years ago
Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocke
umka2103 [35]

Answer:

v_{f} = 115.95 m / s

Explanation:

This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions

        Thrust = v_{e}  \frac{dM}{dt}

        v_{f}-v₀ = v_{e} ln ( \frac{M_{o} }{M_{f}} )

where v_{e} is the velocity of the gases relative to the rocket

let's apply these expressions to our case

the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units

       M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg

     

The final mass is the mass of the engines + the mass of the rocket

      M_{f} = 25.5 +54.5 = 80 g = 0.080 kg

thrust and duration of ignition are given

       thrust = 5.26 N

       t = 1.90 s

Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear

          thrust = v_{e} \frac{M_{f} - M_{o}  }{t_{f} - t_{o}  }

          v_{e} = thrust  \frac{\Delta t}{\Delta M}

          v_{e} = 5.26 \frac{1.90}{0.080 -0.0927}

          v_{e} = - 786.93 m / s

the negative sign indicates that the direction of the gases is opposite to the direction of the rocket

now we look for the final speed of the rocket, which as part of rest its initial speed is zero

            v_{f}-0 = v_{e} ln ( \frac{M_{o} }{M_{f} } )

we calculate

            v_{f} = 786.93 ln (0.0927 / 0.080)

            v_{f} = 115.95 m / s

5 0
3 years ago
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D. sound energy........

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