Hey there!:
Here the Statement - D is correct.
Because Orbitals containing the core electrons are more attracted towards nuclear charge and hence less shilded from nuclear charge than an orbital that doesn't penetrate. Also due to more attraction between the orbital containing core electron and nucleus, it will have less energy.
Hope this helps!
KE=1/2 m v^2
KE= .5 x 2kg x 15m/s to the 2nd power
KE=225 km/s
Answer:
281 K
Explanation:
Charles's Law. V1/T1 = V2/T2.
The temperature must be in K = 21.6°C + 273 = 294.6K.
V1T2 = V2T1.
3.62L x T2 = 3.45L x 294.6K
T2 = (3.45 x 294.6) / 3.62 = 1016.4 / 3.62 = B): 281K.
(By direct proportion of volume change: (3.45L / 3.62L) x 294.6K = 281K).
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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