Answer:
A 70 kg box is slid along the floor by a 400 n force. The coefficient of friction between the box and the floor is 0. 50 when the box is sliding
Answer:
Temperature at the exit = 
Explanation:
For the steady energy flow through a control volume, the power output is given as
Inlet area of the turbine = 
To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.
Assuming Argon behaves as an Ideal gas, we have the specific volume 
as
for Ideal gasses, the enthalpy change can be calculated using the formula
hence we have
<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>
evaluating the above equation, we have 
Hence, the temperature at the exit =
Answer:
no the moon does not rotate it only goes in circle just like the sun so I disagree with your friend
Answer:
ΔE = 37.8 x 10^9 J
Explanation:
The energy required will increased the potential energy and increase the kinetic energy.
As the altitude change is fairly small compared to the earth radius, we can ASSUME that the average gravity will be a good representative
Gravity acceleration at altitude would be 9.8(6400²/8000²) = 6.272 m/s²
G(avg) = (9.8 + 6.272)/2 = 8.036 m/s²
ΔPE = mG(avg)Δh = 1000(8.036)(8e6 - 6.4e6) = 12.857e9 J
The centripetal force at orbit must be equal to the gravity force
mv²/R = mg'
v²/8.0e6 = 6.272
v² = (6.272(8.0e6)) = 50.2e6 m²/s²
The maximum velocity when resting on earth at the equator is about 460 m/s.
The change in kinetic energy is
ΔKE = ½m(vf² - vi²)(1000)
ΔKE = ½(1000)(50.2e6 - 460²) = 25e9 J
Total energy increase is
25e9 + 12.857e9 = 37.8e9 J
Explanation:
Period P has units of seconds (s).
Length has units of meters (m).
Mass has units of kilograms (kg).
Acceleration has units of meters per second squared (m/s²).
Dimensional analysis:
s = √(m / (m/s²))
Therefore:
P = k √(L/g)
where k is a dimensionless constant.
