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>
Answer:
Resistance, 
Explanation:
Given that,
Voltage of the battery, V = 9 volts
Current produced in the circuit, I = 17 A
We need to find the resistance when shorted by a wire of negligible resistance. It is a case of Ohm's law. The voltage is given by :
So, the resistance in the circuit is 0.529 ohms. Hence, this is the required solution.
Answer:
3 seconds
Explanation:
Applying,
Applying,
v = u±gt................ Equation 1
Where v = final velocity, u = initial velocity, t = time, g = acceleration due to gravity.
From the question,
Given: v = 0 m/s ( at the maximum height), u = 30 m/s
Constant: g = -10 m/s
Substitute these values into equation 1
0 = 30-10t
10t = 30
t = 30/10
t = 3 seconds
Answer:
24 Coulumbs
Explanation:
Given data
time= 1 minute= 6 seconds
P=2 W
R= 12 ohm
We know that
P= I^2R
P/R= I^2
2/12= I^2
I^2= 0.166
I= √0.166
I= 0.4 amps
We know also that
Q= It
substitute
Q= 0.4*60
Q= 24 Columbs
Hence the charge is 24 Coulumbs
