Answer: C) a natural resource that humans use to generate energy
So basically its anything found on earth to make energy. This includes nonrenewable sources such as fossil fuels (coal or natural gas), and renewable sources such as wind, hydro or solar.
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 hypothesis is an educated guess, while a theory is usually a tested, well substained explanation of something.
Answer: 40 N
Explanation: Use the formula
F = m a
to solve this.
F = (8 kg) (5 m/s²)
F = 40 kgm/s²
F = 40 N
Answer:
Option 5:
The 60W bulb has a greater resistance and a lower current than the 100 W bulb.
Explanation:
We have to compare the resistance and current of both bulbs.
Bulb A
Power = 60 W
Voltage = 110 V
Power is given as:
![P = V^2 /R](https://tex.z-dn.net/?f=P%20%3D%20V%5E2%20%2FR)
where V= voltage and R = resistance
![=> 60 = 110^2 / R\\\\R = 201.6 \Omega](https://tex.z-dn.net/?f=%3D%3E%2060%20%3D%20110%5E2%20%2F%20R%5C%5C%5C%5CR%20%3D%20201.6%20%5COmega)
Power is also given as:
P = IV
where I = current
=> 60 = I * 110
I = 60/110 = 0.54 A
Bulb B
Power = 100 W
Voltage = 110 V
To get resistance:
![100 = 110^2 / R\\\\R = 121 \Omega](https://tex.z-dn.net/?f=100%20%3D%20110%5E2%20%2F%20R%5C%5C%5C%5CR%20%3D%20121%20%5COmega)
To get current:
100 = I * 110
I = 100 / 110
I = 0.91 A
Therefore, by comparison, the 60W bulb has a greater resistance and a lower current.