B. Gabriella is slowing down at the same rate that Kendall is speeding up, and Franklin is not accelerating.
Answer:
distance = 6.1022 x10^16[m]
Explanation:
To solve this problem we must use the formula of the average speed which relates distance to time, so we have
v = distance / time
where:
v = velocity = 3 x 10^8 [m/s]
distance = x [meters]
time = 6.45 [light years]
Now we have to convert from light-years to seconds in order to get the distance in meters.
![t = 6.45 [light-years]*365[\frac{days}{1light-year}]*24[\frac{hr}{1day}] *60[\frac{min}{1hr}]*60[\frac{seg}{1min} ] =203407200 [s]](https://tex.z-dn.net/?f=t%20%3D%206.45%20%5Blight-years%5D%2A365%5B%5Cfrac%7Bdays%7D%7B1light-year%7D%5D%2A24%5B%5Cfrac%7Bhr%7D%7B1day%7D%5D%20%2A60%5B%5Cfrac%7Bmin%7D%7B1hr%7D%5D%2A60%5B%5Cfrac%7Bseg%7D%7B1min%7D%20%5D%20%3D203407200%20%5Bs%5D)
Now using the formula:
distance = v * time
distance = (3*10^8)*203407200
distance = 6.1022 x10^16[m]
The power that heat pump draws when running will be 6.55 kj/kg
A heat pump is a device that uses the refrigeration cycle to transfer thermal energy from the outside to heat a building (or a portion of a structure).
Given a heat pump used to heat a house runs about one-third of the time. The house is losing heat at an average rate of 22,000 kJ/h and if the COP of the heat pump is 2.8
We have to determine the power the heat pump draws when running.
To solve this question we have to assume that the heat pump is at steady state
Let,
Q₁ = 22000 kj/kg
COP = 2.8
Since heat pump used to heat a house runs about one-third of the time.
So,
Q₁ = 3(22000) = 66000 kj/kg
We known the formula for cop of heat pump which is as follow:
COP = Q₁/ω
2.8 = 66000 / ω
ω = 66000 / 2.8
ω = 6.66 kj/kg
Hence the power that heat pump draws when running will be 6.55 kj/kg
Learn more about heat pump here :
brainly.com/question/1042914
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True the magnitude is the velocity of speed
Answer:
C
Explanation:
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