Four regions of the electromagnetic spectrum that astronomers use when observing objects in the space are the following enumerated answers.
1. First is Ultraviolet
2. Next is Infrared
3. Then the radio
4. Lastly the Visible lights.
These are the answers to the problem.
Answer:
![P_1 = 1166.7 Watt](https://tex.z-dn.net/?f=P_1%20%3D%201166.7%20Watt)
![P_2 = 2000 Watt](https://tex.z-dn.net/?f=P_2%20%3D%202000%20Watt)
Explanation:
Average power for the human sprinter is given as
![Power = \frac{\Delta E}{\Delta t}](https://tex.z-dn.net/?f=Power%20%3D%20%5Cfrac%7B%5CDelta%20E%7D%7B%5CDelta%20t%7D)
so we have
![P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20-%200%7D%7B%5CDelta%20t%7D)
![P = \frac{\frac{1}{2}(70)(10^2) - 0}{3}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D%2870%29%2810%5E2%29%20-%200%7D%7B3%7D)
![P_1 = 1166.7 Watt](https://tex.z-dn.net/?f=P_1%20%3D%201166.7%20Watt)
Average power for greyhound is given as
![P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20-%200%7D%7B%5CDelta%20t%7D)
![P = \frac{\frac{1}{2}(30)(20^2) - 0}{3}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D%2830%29%2820%5E2%29%20-%200%7D%7B3%7D)
![P_2 = 2000 Watt](https://tex.z-dn.net/?f=P_2%20%3D%202000%20Watt)
Answer:
229,098.96 J
Explanation:
mass of water (m) = 456 g = 0.456 kg
initial temperature (T) = 25 degrees
final temperature (t) = - 10 degrees
specific heat of ice = 2090 J/kg
latent heat of fusion =33.5 x 10^(4) J/kg
specific heat of water = 4186 J/kg
for the water to be converted to ice it must undergo three stages:
- the water must cool from 25 degrees to 0 degrees, and the heat removed would be Q = m x specific heat of water x change in temp
Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J
- the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp
Q = 0.456 x 33.5 x 10^(4) = 152760 J
- the water must cool further to -10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp
Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J
The quantity of heat removed from all three stages would be added to get the total heat removed.
Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J
550! OBVY! lol! ope this helps1