Answer:
m v^2 / R = G M m / R^2 gravitational attraction = centripetal force
M = v^2 R / G solving for M
period = 6 h 25 min = (6 * 3600 + 25 * 60) sec = 23,100 sec = T
v = 2 pi R / T
M = 4 pi^2 R^3 / (G T^2)
M = 39.5 * (8.6E7)^3 / (6.67E-11 * 2.31E4^2)
M = 39.5 * 636 / (6.67 * 5.34) * 10^24
M = 7.05 * 10^26 kg
Sally's average speed is <u>35.3 mi/h.</u>
Average speed of a body is the total distance traveled in the given time interval.
Express the distance d traveled in miles.

Express the time t traveled in hours.

Calculate the average speed v.

Her average speed is 35.3 mi/h, which is less than the speed limit of 65 mi/h.
However, the average speed of an object is different from its instantaneous speed. It could be possible that at the time when the officer apprehended her, Sally could have been travelling at a speed greater than the prescribed speed limit, which would have prompted the officer to issue a speeding ticket to her.
Thus, the average speed of a person cannot be considered as a bench mark for speeding offences, since her instantaneous speed could have been higher than the speeding limit and yet she could have had an average speed less than the speeding limit.
Answer:
T = 27.92 N
Explanation:
For this exercise let's use Newton's second law
T - W = m a
The weight
W = mg
The acceleration can be found by derivatives
a = dv / dt
v = 2 t + 0.6 t²
a = 2 + 0.6 t
We replace
T - mg = m (2 + 0.6t)
T = m (g + 2 + 0.6 t) (1)
Let's look for the time for the speed of 15 m / s
15 = 2 t + 0.6 t²
0.6 t² + 2 t - 15 = 0
We solve the second degree equation
t = [-2 ±√(4 - 4 0.6 (-15))] / 2 0.6
t = [-2 ±√40] / 1.3 = [-2 ± 6.325] / 1.2
We take the positive time
t = 3.6 s
Let's calculate from equation 1
T = 2.00 (9.8 + 2 + 0. 6 3.6)
T = 27.92 N
Answer:
See explanation
Explanation:
Notice that the condenser section includes both the hot water and space heater and station (3) is specified as being in the Quality region. Assume that 50°C is a reasonable maximum hot water temperature for home usage, thus at a high pressure of 1.6 MPa, the maximum power available for hot water heating will occur when the refrigerant at station (3) reaches the saturated liquid state. (Quick Quiz: justify this statement). Assume also that the refrigerant at station (4) reaches a subcooled liquid temperature of 20°C while heating the air.
Using the conditions shown on the diagram and assuming that station (3) is at the saturated liquid state
a) On the P-h diagram provided below carefully plot the five processes of the heat pump together with the following constant temperature lines: 50°C (hot water), 13°C (ground loop), and -10°C (outside air temperature)
b) Using the R134a property tables determine the enthalpies at all five stations and verify and indicate their values on the P-h diagram.
c) Determine the mass flow rate of the refrigerant R134a. [0.0127 kg/s]
d) Determine the power absorbed by the hot water heater [2.0 kW] and that absorbed by the space heater [0.72 kW].
e) Determine the time taken for 100 liters of water at an initial temperature of 20°C to reach the required hot water temperature of 50°C [105 minutes].
f) Determine the Coefficient of Performance of the hot water heater [COPHW = 4.0] (defined as the heat absorbed by the hot water divided by the work done on the compressor)
g) Determine the Coefficient of Performance of the heat pump [COPHP = 5.4] (defined as the total heat rejected by the refrigerant in the hot water and space heaters divided by the work done on the compressor)
h) What changes would be required of the system parameters if no geothermal water loop was used, and the evaporator was required to absorb its heat from the outside air at -10°C. Discuss the advantages of the geothermal heat pump system over other means of space and water heating