Yes, 5 ºC = 278 K
278 yards is approximately 254 meters.
We have to make a few assumptions first. We assume the angle provided (12.9 degrees) is the plane's attitude, that is, the angle with respect to the horizontal. We also assume the plane's nose is angled toward the ground because you asked for rate of descent (not dissent). We also assume that all of the plane's airspeed is in the same direction as its attitude (which is often not the case in actual flight, where an aircraft's velocity vector may have a different angle than the angle the plane makes with the ground). So, we have an airplane that's traveling at 278 km per hour at an angle that is -12.9 degrees to the horizontal. (The minus sign indicates a downward direction.) That describes a vector having vertical and horizontal components. We are interested in the vertical component, which is given by the formula Vy = Vsin(theta) = 278sin(-12.9) = -62.1. The rate of descent, therefore, is 62.1 km/hr. It is not necessary to use the minus sign if you say "rate of descent," because the minus sign is implied. If you are asked what the vertical velocity is, then you must use the minus sign to indicate whether it's positive (ascending) or negative (descending).
The lowest temperature every recorded in Honolulu was 52 degrees. Honolulu is a city with pleasant weather and has 278 days of sunshine each year.
To find the average velocity of atoms in neon at 278 K, we can use the equation for the root mean square speed (v_rms) given by (v_{rms} = \sqrt{\frac{3kT}{m}}), where (k) is the Boltzmann constant ((1.38 \times 10^{-23} , \text{J/K})), (T) is the temperature in Kelvin, and (m) is the mass of a neon atom in kilograms. The molar mass of neon is approximately 20.18 g/mol, which converts to (3.34 \times 10^{-26} , \text{kg}) per atom. Plugging in the values, the average velocity of neon atoms at 278 K is approximately 394 m/s.
3,000 square feet equates to 278 (278.7091) square meters.
Kelvin = Celsius + 273 278 Kelvin = 5 Celsius
5 degrees Celsius is equal to 41 degrees Fahrenheit.
It is very close. To be exact, 278.15 K = 5 C
41oF5C = 41Fis C 278 K5° Celsius is equal to 41° Fahrenheit.To convert °C to °F multiply the Celsius temperature by 9, divide the sum by 5, and add 32.
Start by taking the number in Fahrenheit and subtracting 32. Then divide the number by 9, and then multiply it by 5. This is how you convert Fahrenheit to Celsius or use the equation C = (F - 32) × 5/9In this case, the answer is about 1,371.11 degrees Celsius.
Celsius+273=Kelvin ex. 5 Celsius = 278 Kelvin
kelvin would change from 308 to 278
Since absolute zero is -273.15 °C and 0 °K, 278 K approximately the same as 5 °C. On the Fahrenheit scale, it would be equal to 41 °F.
It is where no more heat can be lost. Theoretically when all molecular activity is lost. It is used when comparisons of energy are needed. You cant use Celsius for that since you're starting out at 278 degrees. The values wouldn't be true percentages.
75% of 278= 75% * 278= 0.75 * 278= 208.5
55% of 278= 55% * 278= 0.55 * 278= 152.9
15% of 278= 15% * 278= 0.15 * 278= 41.7