Answer: It is both B and D
Select all that apply.
At night, thermal energy moves _____.
from space to the atmosphere
from the land to the atmosphere
from the atmosphere to the land
from the atmosphere to space
Answer:
okay sooo the weight is: 294 n
the normal force is 286 n
the acceleration is: -0.38 m/s²
Answer:
v= 0.2 m/s
Explanation:
Given that
m₁ = 50 kg
m₂ = 100 g = 0.1 kg
u =10 0 m/s
If there is no any external force on the system then the total linear momentum of the system will be conserve.
Initial linear momentum = Final momentum
m₁u₁ + m₂ u₂ =m₂ v₂ +m₁v₁
50 x 0 + 0.1 x 100 = 50 v + 0
0+ 10 = 50 v
v= 0.2 m/s
Therefore the recoil speed will be 0.2 m/s.
Answer:
V= 33.98 m/s
Explanation:
Given that
Horizontal speed ,u= 17 m/s
Time taken by rockets to strike the water ,t= 3 s
We know that acceleration due to gravity ,g= 9.81 m/s²
There is no any acceleration in the horizontal direction that is why the horizontal veloity will remain constant.
In the vertical direction
vy = uy+ g t
Initial velocity in vertical direction is 0 m/s.
vy= 0+ 9.81 x 3
vy = 29.43 m/s
The resultant velocity
V= 33.98 m/s
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
#SPJ4
