Answer:
38.6 N
2.57 m/s²
Explanation:
Draw a free body diagram of the box. There are four forces:
Weight force mg pulling down,
Normal force N pushing up,
Friction force Nμ pushing left,
and applied force P pulling at an angle 40°.
Sum of forces in the y direction:
∑F = ma
N + P sin 40° − mg = 0
N = mg − P sin 40°
The net force in the x direction is:
∑F = P cos 40° − Nμ
∑F = P cos 40° − (mg − P sin 40°) μ
∑F = P cos 40° − mgμ + Pμ sin 40°
∑F = P (cos 40° + μ sin 40°) − mgμ
Plugging in values:
∑F = (80 N) (cos 40° + 0.23 sin 40°) − (15 kg) (10 m/s²) (0.23)
∑F = 38.6 N
Net force equals mass times acceleration:
∑F = ma
38.6 N = (15 kg) a
a = 2.57 m/s²
Large amounts of water do have a big impact on the weather: indeed, it takes less energy to warm/cool land than water.
Therefore, places near large amounts of water tend to have smaller differences in temperature between summer and winter than places far from waters.
Hence, during winter in Puerto Rico, alongside the coast, the temperature will be higher than in the innermost parts of the island.
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
Nuclear fusion because atomic nuclei combine to form a heavier nucleus. Option A is correct.
<h3>What is nuclear fusion?</h3>
The process by which two or more tiny nuclei unite to generate a bigger nucleus is known as a nuclear fusion reaction.
The more energy it takes to liberate an electron from a smaller atom. This is referred to as binding energy.
As a result, when two little nuclei fuse together, there is more binding energy than when two big nuclei fuse together.
For example, the fusion of two hydrogen atoms produces more energy than the fusion of one helium atom, and surplus energy is expelled into space upon binding.
Nuclear fusion because atomic nuclei combine to form a heavier nucleus.
Hence, option A is correct.
To learn more about nuclear fusion refer to the link;
brainly.com/question/14019172
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(a) 5.66 m/s
The flow rate of the water in the pipe is given by

where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have

the radius of the pipe is
r = 0.260 m
So the cross-sectional area is

So we can re-arrange the equation to find the speed of the water:

(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:

where we have

and where
is the cross-sectional area of the pipe at the second point.
Solving for A2,

And finally we can find the radius of the pipe at that point:
