It can be inferred from the given table that the ratio of voltage with current is a constant, which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys Ohm's law. According to Ohm's law, the ratio of voltage with current is the resistance of the conductor. Hence, the resistRead more
It can be inferred from the given table that the ratio of voltage with current is a constant, which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys Ohm’s law. According to Ohm’s law, the ratio of voltage with current is the resistance of the conductor. Hence, the resistance of manganin is 19.7 Ω .
Resistivity of aluminium, ρAl = 2.63 x 10-8 Ω m Relative density of aluminium, d₁= 2.7 Let l₁ be the length of aluminium wire and m₁ be its mass. Resistance of the aluminium wire = R₁ Area of cross-section of the aluminium wire = A₁ Resistivity of copper, ρcu = 1.72 x 10-8 Ωm Relative density of copRead more
Resistivity of aluminium, ρAl = 2.63 x 10-8 Ω m
Relative density of aluminium, d₁= 2.7
Let l₁ be the length of aluminium wire and m₁ be its mass.
Resistance of the aluminium wire = R₁
Area of cross-section of the aluminium wire = A₁
Resistivity of copper, ρcu = 1.72 x 10-8 Ωm
Relative density of copper, d₂ = 8.9
Let l₂ be the length of copper wire and m₂ be its mass.
Resistance of the copper wire = R2
Area of cross-section of the copper wire = A2
The two relations can be written as
R₁ =ρ₁ l₁/A₁ ——————–Eq-1
R2 =ρ2 l2A2 ——————–Eq-2
It is given that ,R₁=R2 => ρ₁ l₁/A₁= ρ2 l2A2
=> ρ₁ /A₁ = ρ2 /A2 ( as ,l₁=l2)
=> A₁ /A2 =ρ₁/ρ2 = (2.63 x 10⁻⁸ )/(1.72 x 10⁻⁸ )=2.63/1.72
Mass of the copper wire ,m₁= Volume x Density
=A₁L₁ x d₁ =A₁l₁d₁———————-Eq -3
Mass of the copper wire ,m2= Volume x Density
=A2L2 x d2 =A2l2d2———————-Eq -4,from which we obtain
m₁/m2 = (A1l1d1)/(A2l2d2)
For l1=l2
m1/m2 = A1d1/A2d2
For A1/A2 =2.63/1.72,
Therefore
m1/m2= 2.63/1.72 x 2.7/8.9 = 0.46
It can be inferred from this ratio that m1 is less than m2. Hence, aluminium is lighter than copper. Since aluminium is lighter, it is preferred for overhead power cables over copper.
Ans (a). Number of secondary cells, n = 6 Emf of each secondary cell, E = 2.0 V Internal resistance of each cell, r = 0.015 Ω series resistor is connected to the combination of cells. Resistance of the resistor, R = 8.5 Ω Current drawn from the supply = I, which is given by the relation, I = (nE)/(RRead more
Ans (a).
Number of secondary cells, n = 6
Emf of each secondary cell, E = 2.0 V
Internal resistance of each cell, r = 0.015 Ω series resistor is connected to the combination of cells. Resistance of the resistor, R = 8.5 Ω
Current drawn from the supply = I, which is given by the relation,
I = (nE)/(R+nr)
= (6×2)/(8.5 + 6 x 0.015) = 12 /8.59 =1.39 A
Terminal voltage, V = IR = 1.39 x 8.5 = 11.87 A
Therefore, the current drawn from the supply is 1.39 A and terminal voltage is 11.87 A.
Ans (b).
After a long use, emf of the secondary cell, E = 1.9 V
Internal resistance of the cell, r = 380 Ω
Hence, maximum current =E/r=1.9/380 = 0.005 A
Therefore, the maximum current drawn from the cell is 0.005 A. Since a large current is required to start the motor of a car, the cell cannot be used to start a motor.
Number density of free electrons in a copper conductor, n = 8.5 x 1028 m-3 Length of the copper wire, I = 3.0 m Area of cross-section of the wire, A = 2.0 x 10-6 m2 Current carried by the wire, I = 3.0 A, which is given by the relation, I = nAeVd Where, e = Electric charge = 1.6 x 10-19 C Vd= DriftRead more
Number density of free electrons in a copper conductor, n = 8.5 x 1028 m–3 Length of the copper wire, I = 3.0 m
Area of cross-section of the wire, A = 2.0 x 10-6 m2
Current carried by the wire, I = 3.0 A, which is given by the relation, I = nAeVd Where, e = Electric charge = 1.6 x 10-19 C
Vd= Drift Velocity = Length of the wire (l) /Time taken to cover l (t)
I = nAel/t => t = nAel/t
t = 3x 8.5 x 1028 x 2 x 10-6 x 1.6 x 10-19 )/3.0
= 2.7 x 10⁴ s
Therefore, the time taken by an electron to drift from one end of the wire to the other is 2.7 x 104 s.
Emf of the cell, E₁ = 1.25 V Balance point of the potentiometer, l₁= 35 cm. The cell is replaced by another cell of emf E2. New balance point of the potentiometer, l2 = 63 cm The balance condition is given by the relation. E₁/E2= l₁/l2 =>E2 = = E₁l2/l₁ = 1.25 x 63/35 =2.25 V Therefore, emf ofRead more
Emf of the cell, E₁ = 1.25 V
Balance point of the potentiometer, l₁= 35 cm. The cell is replaced by another cell of emf E2.
New balance point of the potentiometer, l2 = 63 cm The balance condition is given by the relation.
Emf of the storage battery, E = 8.0 V Internal resistance of the battery, r = 0.5 Ω DC supply voltage, V = 120 V Resistance of the resistor, R = 15.5 Ω Effective voltage in the circuit = V1 R is connected to the storage battery in series. Hence, it can be written as V1 = V - E V1 = 120 - 8 = 112 V CRead more
Emf of the storage battery, E = 8.0 V
Internal resistance of the battery, r = 0.5 Ω
DC supply voltage, V = 120 V
Resistance of the resistor, R = 15.5 Ω
Effective voltage in the circuit = V1
R is connected to the storage battery in series. Hence, it can be written as V1 = V – E
V1 = 120 – 8 = 112 V
Current flowing in the circuit = I, which is given by the relation,
I =V1/(R+r) = 112/(15.5 +5) = 112/16 =7 A
Voltage across resistor R given by the product, IR = 7 x 15.5 = 108.5 V
DC supply voltage = Terminal voltage of battery + Voltage drop across R Terminal voltage of battery = 120 – 108.5 = 11.5 V
A series resistor in a charging circuit limits the current drawn from the external source.
The current will be extremely high in its absence. This is very dangerous.
Supply voltage, V = 230 V Initial current drawn, I₁ = 3.2 A Initial resistance = R₁, which is given by the relation, R₁=V/I =230 /3.2 =71.87 Ω Steady state value of the current, I2 = 2.8 A Resistance at the steady state = R2, which is given as R2=230/2.8 = 82.14 Ω Temperature co-efficient of nichromRead more
Supply voltage, V = 230 V
Initial current drawn, I₁ = 3.2 A
Initial resistance = R₁, which is given by the relation,
R₁=V/I =230 /3.2 =71.87 Ω
Steady state value of the current, I2 = 2.8 A
Resistance at the steady state = R2, which is given as
R2=230/2.8 = 82.14 Ω
Temperature co-efficient of nichrome,α = 1.70 x 10⁻4 oC⁻1
Initial temperature of nichrome, T₁= 27.0°C
Study state temperature reached by nichrome = T2
T2can be obtained by the relation for α,
α =(R2-R₁)/R ₁(T2-T₁)
(T2-T₁) =(R2-R₁)/α
=> T2-27 = (82.14-71.87 ) /(71.87 x 1.7 x 10⁻4) = 840.5
=> T2 =840.5 + 27 =867.5 °C
Therefore, the steady temperature of the heating element is 867.5°C
Temperature, T₁ = 27.5°C Resistance of the silver wire at T₁, R₁ = 2.1 Ω Temperature, T2 = 100°C Resistance of the silver wire at T2, R2 = 2.7 Ω Temperature coefficient of silver = α It is related with temperature and resistance as α =(R2-R₁)/R ₁(T2-T₁) = (2.7 -2.1 ) /2.1 (100 -27.5) = 0.0039 °C⁻¹Read more
Temperature, T₁ = 27.5°C
Resistance of the silver wire at T₁, R₁ = 2.1 Ω
Temperature, T2 = 100°C
Resistance of the silver wire at T2, R2 = 2.7 Ω
Temperature coefficient of silver = α
It is related with temperature and resistance as
α =(R2-R₁)/R ₁(T2-T₁)
= (2.7 -2.1 ) /2.1 (100 -27.5) = 0.0039 °C⁻¹
Therefore, the temperature coefficient of silver is 0.0039°C⁻¹.
Length of the wire, l =15 m Area of cross-section of the wire, a = 6.0 x 10⁻7 m2, Resistance of the material of the wire, R = 5.0Ω. Resistivity of the material of the wire = ρ, Resistance is related with the resistivity as R = ρ l /A => ρ = RA/l = 5 x 6.0 x 10⁻7 / 15 = 2x 10⁻7 Ωm Therefore, the rRead more
Length of the wire, l =15 m
Area of cross-section of the wire, a = 6.0 x 10⁻7 m2, Resistance of the material of the wire, R = 5.0Ω. Resistivity of the material of the wire = ρ, Resistance is related with the resistivity as
R = ρ l /A
=> ρ = RA/l
= 5 x 6.0 x 10⁻7 / 15 = 2x 10⁻7 Ωm
Therefore, the resistivity of the material is 2 x 10_7Ωm.
Ans (a). We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential. Ans (b). Yes, the man will get an electric shock if he touches the metal slab next morning. The steady dischargRead more
Ans (a).
We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.
Ans (b).
Yes, the man will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminum sheet. As a result, its voltage rises gradually. The rise in the voltage depends on the capacitance of the capacitor formed by the aluminum slab and the ground.
Ans (c).
The occurrence of thunderstorms and lightning charges the atmosphere continuously. Hence, even with the presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral.
Ans (d).
During lightning and thunderstorm, light energy, heat energy, and sound energy are dissipated in the atmosphere.
What conclusion can you draw from the following observations on a resistor made of alloy manganin?
It can be inferred from the given table that the ratio of voltage with current is a constant, which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys Ohm's law. According to Ohm's law, the ratio of voltage with current is the resistance of the conductor. Hence, the resistRead more
It can be inferred from the given table that the ratio of voltage with current is a constant, which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys Ohm’s law. According to Ohm’s law, the ratio of voltage with current is the resistance of the conductor. Hence, the resistance of manganin is 19.7 Ω .
See lessTwo wires of equal length, one of aluminum and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminum wires are preferred for overhead power cables. (ρAl = 2.63 × 10⁻⁸ Ω m, ρCu = 1.72 × 10⁻⁸Ω m, Relative density of Al = 2.7, of Cu = 8.9.)
Resistivity of aluminium, ρAl = 2.63 x 10-8 Ω m Relative density of aluminium, d₁= 2.7 Let l₁ be the length of aluminium wire and m₁ be its mass. Resistance of the aluminium wire = R₁ Area of cross-section of the aluminium wire = A₁ Resistivity of copper, ρcu = 1.72 x 10-8 Ωm Relative density of copRead more
Resistivity of aluminium, ρAl = 2.63 x 10-8 Ω m
Relative density of aluminium, d₁= 2.7
Let l₁ be the length of aluminium wire and m₁ be its mass.
Resistance of the aluminium wire = R₁
Area of cross-section of the aluminium wire = A₁
Resistivity of copper, ρcu = 1.72 x 10-8 Ωm
Relative density of copper, d₂ = 8.9
Let l₂ be the length of copper wire and m₂ be its mass.
Resistance of the copper wire = R2
Area of cross-section of the copper wire = A2
The two relations can be written as
R₁ =ρ₁ l₁/A₁ ——————–Eq-1
R2 =ρ2 l2A2 ——————–Eq-2
It is given that ,R₁=R2 => ρ₁ l₁/A₁= ρ2 l2A2
=> ρ₁ /A₁ = ρ2 /A2 ( as ,l₁=l2)
=> A₁ /A2 =ρ₁/ρ2 = (2.63 x 10⁻⁸ )/(1.72 x 10⁻⁸ )=2.63/1.72
Mass of the copper wire ,m₁= Volume x Density
=A₁L₁ x d₁ =A₁l₁d₁———————-Eq -3
Mass of the copper wire ,m2= Volume x Density
=A2L2 x d2 =A2l2d2———————-Eq -4,from which we obtain
m₁/m2 = (A1l1d1)/(A2l2d2)
For l1=l2
m1/m2 = A1d1/A2d2
For A1/A2 =2.63/1.72,
Therefore
m1/m2= 2.63/1.72 x 2.7/8.9 = 0.46
It can be inferred from this ratio that m1 is less than m2. Hence, aluminium is lighter than copper. Since aluminium is lighter, it is preferred for overhead power cables over copper.
See less(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage? (b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?
Ans (a). Number of secondary cells, n = 6 Emf of each secondary cell, E = 2.0 V Internal resistance of each cell, r = 0.015 Ω series resistor is connected to the combination of cells. Resistance of the resistor, R = 8.5 Ω Current drawn from the supply = I, which is given by the relation, I = (nE)/(RRead more
Ans (a).
Number of secondary cells, n = 6
Emf of each secondary cell, E = 2.0 V
Internal resistance of each cell, r = 0.015 Ω series resistor is connected to the combination of cells. Resistance of the resistor, R = 8.5 Ω
Current drawn from the supply = I, which is given by the relation,
I = (nE)/(R+nr)
= (6×2)/(8.5 + 6 x 0.015) = 12 /8.59 =1.39 A
Terminal voltage, V = IR = 1.39 x 8.5 = 11.87 A
Therefore, the current drawn from the supply is 1.39 A and terminal voltage is 11.87 A.
Ans (b).
After a long use, emf of the secondary cell, E = 1.9 V
Internal resistance of the cell, r = 380 Ω
Hence, maximum current =E/r=1.9/380 = 0.005 A
Therefore, the maximum current drawn from the cell is 0.005 A. Since a large current is required to start the motor of a car, the cell cannot be used to start a motor.
See lessThe number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10²⁸ m⁻³. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10⁻⁶ m² and it is carrying a current of 3.0 A.
Number density of free electrons in a copper conductor, n = 8.5 x 1028 m-3 Length of the copper wire, I = 3.0 m Area of cross-section of the wire, A = 2.0 x 10-6 m2 Current carried by the wire, I = 3.0 A, which is given by the relation, I = nAeVd Where, e = Electric charge = 1.6 x 10-19 C Vd= DriftRead more
Number density of free electrons in a copper conductor, n = 8.5 x 1028 m–3 Length of the copper wire, I = 3.0 m
Area of cross-section of the wire, A = 2.0 x 10-6 m2
Current carried by the wire, I = 3.0 A, which is given by the relation, I = nAeVd Where, e = Electric charge = 1.6 x 10-19 C
Vd= Drift Velocity = Length of the wire (l) /Time taken to cover l (t)
I = nAel/t => t = nAel/t
t = 3x 8.5 x 1028 x 2 x 10-6 x 1.6 x 10-19 )/3.0
= 2.7 x 10⁴ s
Therefore, the time taken by an electron to drift from one end of the wire to the other is 2.7 x 104 s.
See lessIn a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?
Emf of the cell, E₁ = 1.25 V Balance point of the potentiometer, l₁= 35 cm. The cell is replaced by another cell of emf E2. New balance point of the potentiometer, l2 = 63 cm The balance condition is given by the relation. E₁/E2= l₁/l2 =>E2 = = E₁l2/l₁ = 1.25 x 63/35 =2.25 V Therefore, emf ofRead more
Emf of the cell, E₁ = 1.25 V
Balance point of the potentiometer, l₁= 35 cm. The cell is replaced by another cell of emf E2.
New balance point of the potentiometer, l2 = 63 cm The balance condition is given by the relation.
E₁/E2= l₁/l2
=>E2 =
= E₁l2/l₁ = 1.25 x 63/35 =2.25 V
Therefore, emf of the second cell is 2.25V.
See lessA storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
Emf of the storage battery, E = 8.0 V Internal resistance of the battery, r = 0.5 Ω DC supply voltage, V = 120 V Resistance of the resistor, R = 15.5 Ω Effective voltage in the circuit = V1 R is connected to the storage battery in series. Hence, it can be written as V1 = V - E V1 = 120 - 8 = 112 V CRead more
Emf of the storage battery, E = 8.0 V
Internal resistance of the battery, r = 0.5 Ω
DC supply voltage, V = 120 V
Resistance of the resistor, R = 15.5 Ω
Effective voltage in the circuit = V1
R is connected to the storage battery in series. Hence, it can be written as V1 = V – E
V1 = 120 – 8 = 112 V
Current flowing in the circuit = I, which is given by the relation,
I =V1/(R+r) = 112/(15.5 +5) = 112/16 =7 A
Voltage across resistor R given by the product, IR = 7 x 15.5 = 108.5 V
DC supply voltage = Terminal voltage of battery + Voltage drop across R Terminal voltage of battery = 120 – 108.5 = 11.5 V
A series resistor in a charging circuit limits the current drawn from the external source.
The current will be extremely high in its absence. This is very dangerous.
See lessA heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10⁻⁴ °C⁻¹
Supply voltage, V = 230 V Initial current drawn, I₁ = 3.2 A Initial resistance = R₁, which is given by the relation, R₁=V/I =230 /3.2 =71.87 Ω Steady state value of the current, I2 = 2.8 A Resistance at the steady state = R2, which is given as R2=230/2.8 = 82.14 Ω Temperature co-efficient of nichromRead more
Supply voltage, V = 230 V
Initial current drawn, I₁ = 3.2 A
Initial resistance = R₁, which is given by the relation,
R₁=V/I =230 /3.2 =71.87 Ω
Steady state value of the current, I2 = 2.8 A
Resistance at the steady state = R2, which is given as
R2=230/2.8 = 82.14 Ω
Temperature co-efficient of nichrome,α = 1.70 x 10⁻4 oC⁻1
Initial temperature of nichrome, T₁= 27.0°C
Study state temperature reached by nichrome = T2
T2can be obtained by the relation for α,
α =(R2-R₁)/R ₁(T2-T₁)
(T2-T₁) =(R2-R₁)/α
=> T2-27 = (82.14-71.87 ) /(71.87 x 1.7 x 10⁻4) = 840.5
=> T2 =840.5 + 27 =867.5 °C
Therefore, the steady temperature of the heating element is 867.5°C
See lessA silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.
Temperature, T₁ = 27.5°C Resistance of the silver wire at T₁, R₁ = 2.1 Ω Temperature, T2 = 100°C Resistance of the silver wire at T2, R2 = 2.7 Ω Temperature coefficient of silver = α It is related with temperature and resistance as α =(R2-R₁)/R ₁(T2-T₁) = (2.7 -2.1 ) /2.1 (100 -27.5) = 0.0039 °C⁻¹Read more
Temperature, T₁ = 27.5°C
Resistance of the silver wire at T₁, R₁ = 2.1 Ω
Temperature, T2 = 100°C
Resistance of the silver wire at T2, R2 = 2.7 Ω
Temperature coefficient of silver = α
It is related with temperature and resistance as
α =(R2-R₁)/R ₁(T2-T₁)
= (2.7 -2.1 ) /2.1 (100 -27.5) = 0.0039 °C⁻¹
Therefore, the temperature coefficient of silver is 0.0039°C⁻¹.
See lessA negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10⁻⁷ m² , and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?
Length of the wire, l =15 m Area of cross-section of the wire, a = 6.0 x 10⁻7 m2, Resistance of the material of the wire, R = 5.0Ω. Resistivity of the material of the wire = ρ, Resistance is related with the resistivity as R = ρ l /A => ρ = RA/l = 5 x 6.0 x 10⁻7 / 15 = 2x 10⁻7 Ωm Therefore, the rRead more
Length of the wire, l =15 m
Area of cross-section of the wire, a = 6.0 x 10⁻7 m2, Resistance of the material of the wire, R = 5.0Ω. Resistivity of the material of the wire = ρ, Resistance is related with the resistivity as
R = ρ l /A
=> ρ = RA/l
= 5 x 6.0 x 10⁻7 / 15 = 2x 10⁻7 Ωm
Therefore, the resistivity of the material is 2 x 10_7Ωm.
See lessAnswer the following: (a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm⁻¹. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!) (b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m² . Will he get an electric shock if he touches the metal sheet next morning? (c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged? (d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning? (Hint: The earth has an electric field of about 100 Vm⁻¹ at its surface in the downward direction, corresponding to a surface charge density = –10⁻⁹ C m⁻². Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)
Ans (a). We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential. Ans (b). Yes, the man will get an electric shock if he touches the metal slab next morning. The steady dischargRead more
Ans (a).
We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.
Ans (b).
Yes, the man will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminum sheet. As a result, its voltage rises gradually. The rise in the voltage depends on the capacitance of the capacitor formed by the aluminum slab and the ground.
Ans (c).
The occurrence of thunderstorms and lightning charges the atmosphere continuously. Hence, even with the presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral.
Ans (d).
During lightning and thunderstorm, light energy, heat energy, and sound energy are dissipated in the atmosphere.
See less