Answer:
Part a: The electric potential of each sphere is 1.35x10⁸V
Part b: The electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively
Explanation:
As the complete question is not given, the similar question is attached herewith. The values are used as indicated in the given question
Let r_1 = 6 cm=0.06 m
r2 = 2 cm = 0.02 m
Q = 1.2 mC
Let q1 and q2 are the charges on each sphere.
q1 + q2 = 1.2 mC -------(1)
In the equilibrium, V1 = V2
k*q1/r1 = k*q2/r2
q1/0.06 = q2/0.02
q1/q2 = 0.06/0.02
q1/q2 = 3 ---------(2)
On solving equation 1 and 2
we get
q1 = 0.9 mC
q2 = 0.3 mC
So
V1 = k*q1/r1 = (9*10^9*0.9*10^-3)/0.06 = 1.35*10^8 Volts
V2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02 = 1.35*10^8 Volts
So the electric potential of each sphere is 1.35x10⁸V
Part b
Now the electric potential is given as
E1 = k*q1/r1^2 = 9*10^9*0.9*10^-3/0.06^2 = 2.25*10^9 N/C
E2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02^2 = 6.75*10^9 N/C
So the electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively
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:
Explanation:
Brownian motion is a random (irregular) motion of particles e.g smoke particle. The set up in the diagram can be used to observe the motion of smoke.
1. The apparatus used are:
A is a source of light
B is a converging lens
C is a glass smoke cell
D is a microscope
2. The uses of the apparatus are:
A - produces the light required to so as to see clearly the movement of the particles.
B - converges the rays of light from the source to the smoke cell.
C - is made of glass and used for encamping the smoke particles so as not to mix with air.
D - is used for the clear view or observation or study of the motion of the smoke particles in the cell.
Answer:
The potential energy of the rock = 10.5 kN
Explanation:
Mass of rock = 25 kg
Acceleration due to gravity = 10 m/s²
Height = 42 m
Potential energy, PE = mgh, where m is the mass, g is acceleration due to gravity and h is the height.
PE = 25 x 10 x 42 = 10500 N = 10.5 kN
The potential energy of the rock = 10.5 kN
