Answer:
imma need a few answer choices so i can do my research
Explanation:
Please any thank you
Answer:
<em>faster and at a higher luminosity and temperature.</em>
Explanation:
A protostar looks like a star but its core is not yet hot enough for fusion to take place. The luminosity comes exclusively from the heating of the protostar as it contracts. Protostars are usually surrounded by dust, which blocks the light that they emit, so they are difficult to observe in the visible spectrum.
A protostar becomes a main sequence star when its core temperature exceeds 10 million K. This is the temperature needed for hydrogen fusion to operate efficiently.
Stars above about 200 solar masses (Higher mass) generate power so furiously that gravity cannot contain their internal pressure. These stars blow themselves apart and do not exist for long if at all. A protostar with less than 0.08 solar masses never reaches the 10 million K temperature needed for efficient hydrogen fusion. These result in “failed stars” called brown dwarfs which radiate mainly in the infrared and look deep red in color. They are very dim and difficult to detect, but there might be many of them, and in fact they might outnumber other stars in the universe.
That is why higher mass protostars enter the main sequence at a <em>faster and at a higher luminosity and temperature.</em>
A. True. You can use displacement to determine the volume of solids and liquids.
Answer:
True
Explanation:
When a star has reached the main sequence stage, hydrogen is converted into helium by nuclear fusion, . Also, the gravity and pressure are balanced because the star does not radiate more heat than it generates. A star usually spends most of its lifetime at this stage.
The Zero Age Main Sequence is the period during the main sequence when a star stops contracting, and begin to fuse hydrogen in its core.
(a) 5.66 m/s
The flow rate of the water in the pipe is given by
where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have
the radius of the pipe is
r = 0.260 m
So the cross-sectional area is
So we can re-arrange the equation to find the speed of the water:
(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:
where we have
and where 
is the cross-sectional area of the pipe at the second point.
Solving for A2,
And finally we can find the radius of the pipe at that point:
