Answer:
Second drop: 1.04 m
First drop: 1.66 m
Explanation:
Assuming the droplets are not affected by aerodynamic drag.
They are in free fall, affected only by gravity.
I set a frame of reference with the origin at the nozzle and the positive X axis pointing down.
We can use the equation for position under constant acceleration.
X(t) = x0 + v0 * t + 1/2 * a *t^2
x0 = 0
a = 9.81 m/s^2
v0 = 0
Then:
X(t) = 4.9 * t^2
The drop will hit the floor when X(t) = 1.9
1.9 = 4.9 * t^2
t^2 = 1.9 / 4.9
That is the moment when the 4th drop begins falling.
Assuming they fall at constant interval,
Δt = 0.62 / 3 = 0.2 s (approximately)
The second drop will be at:
X2(0.62) = 4.9 * (0.62 - 1*0.2)^2 = 0.86 m
And the third at:
X3(0.62) = 4.9 * (0.62 - 2*0.2)^2 = 0.24 m
The positions are:
1.9 - 0.86 = 1.04 m
1.9 - 0.24 = 1.66 m
above the floor
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
Answer:
a) 1.8°
b) 0.18°
c) 0.018°
Explanation:
Wavelength (λ) = 630nm = 630 *10^-9m
The equation that describes the angular deflection of a dark band is
Wsin(βm) = mλ
w = width of the single slit
λ = wavelength of the light
βm = angular deflection of the mth dark band.
a) In order to get the angular deflection of the first dark band for a slit with 0.02mm width, substitute w = 0.02mm = 0.02*10^-3 , m = 1 , λ= 630*10^-9
0.02*10^-3 sin(β1) = 1 * 630*10^-9
Sin(β1) = 630 * 10^-9 / 0.02*10^-3
Sin(β1) = 0.0315
β1 = Sin^-1(0.0315)
= 1.8°
b) substitute w = 0.2mm = 0.2*10^-3 , m = 1 , λ= 630*10^-9
0.2*10^-3 sin(β1) = 1 * 630*10^-9
Sin(β1) = 630 * 10^-9 / 0.2*10^-3
Sin(β1) = 0.00315
β1 = Sin^-1(0.00315)
= 0.18°
c) substitute w = 2mm = 2*10^-3 , m = 1 , λ= 630*10^-9
2*10^-3 sin(β1) = 1 * 630*10^-9
Sin(β1) = 630 * 10^-9 / 2*10^-3
Sin(β1) = 3.15*10^-4
β1 = Sin^-1(3.15*10^-4)
= 0.018°
The answer is A.
p=m/v
p= 240/60
p= 4 g/cm^3
Because mass does not change from place to place but weight does change from place to place... why? because weight is the amount of gravitational force on an object and mass is the amount of matter in an object. mars has less gravitational force so an object will weigh less than it really weighs there
