In car racing, one advantage of mixing liquid nitrous oxide (N 2 O) with air is that the boiling of the "nitrous" absorbs latent heat of vaporization and thus cools the air and ultimately the fuel-air mixture, allowing more fuel-air mixture to go into each cylinder. As a very rough look at this process, suppose 1.0 mol of nitrous oxide gas at its boiling point, − 88 ℃, is mixed with 4.0 mol of air (assumed diatomic) at 30 ℃. What is the final temperature of the mixture? Use the measured heat capacity of N 2 O at 25 ℃, which is 30.4J/mol ℃. (The primary advantage of nitrous oxide is that it consists of 1/3 oxygen, which is more than air contains, so it supplies more oxygen >to bum the fuel. Another advantage is that its decomposition into nitrogen and oxygen releases energy in the cylinder.)
In car racing, one advantage of mixing liquid nitrous oxide (N 2 O) with air is that the boiling of the "nitrous" absorbs latent heat of vaporization and thus cools the air and ultimately the fuel-air mixture, allowing more fuel-air mixture to go into each cylinder. As a very rough look at this process, suppose 1.0 mol of nitrous oxide gas at its boiling point, − 88 ℃, is mixed with 4.0 mol of air (assumed diatomic) at 30 ℃. What is the final temperature of the mixture? Use the measured heat capacity of N 2 O at 25 ℃, which is 30.4J/mol ℃. (The primary advantage of nitrous oxide is that it consists of 1/3 oxygen, which is more than air contains, so it supplies more oxygen >to bum the fuel. Another advantage is that its decomposition into nitrogen and oxygen releases energy in the cylinder.)
In car racing, one advantage of mixing liquid nitrous oxide (N2O) with air is that the boiling of the "nitrous" absorbs latent heat of vaporization and thus cools the air and ultimately the fuel-air mixture, allowing more fuel-air mixture to go into each cylinder. As a very rough look at this process, suppose 1.0 mol of nitrous oxide gas at its boiling point,
−
88 ℃, is mixed with 4.0 mol of air (assumed diatomic) at 30 ℃. What is the final temperature of the mixture? Use the measured heat capacity of N2O at 25 ℃, which is 30.4J/mol ℃. (The primary advantage of nitrous oxide is that it consists of 1/3 oxygen, which is more than air contains, so it supplies more oxygen >to bum the fuel. Another advantage is that its decomposition into nitrogen and oxygen releases energy in the cylinder.)
4.) The diagram shows the electric field lines of a positively charged conducting sphere of
radius R and charge Q.
A
B
Points A and B are located on the same field line.
A proton is placed at A and released from rest. The magnitude of the work done by the electric field in
moving the proton from A to B is 1.7×10-16 J. Point A is at a distance of 5.0×10-2m from the centre of
the sphere. Point B is at a distance of 1.0×10-1 m from the centre of the sphere.
(a) Explain why the electric potential decreases from A to B. [2]
(b) Draw, on the axes, the variation of electric potential V with distance r from the centre of the
sphere.
R
[2]
(c(i)) Calculate the electric potential difference between points A and B. [1]
(c(ii)) Determine the charge Q of the sphere. [2]
(d) The concept of potential is also used in the context of gravitational fields. Suggest why scientists
developed a common terminology to describe different types of fields. [1]
3.) The graph shows how current I varies with potential difference V across a component X.
904
80-
70-
60-
50-
I/MA
40-
30-
20-
10-
0+
0
0.5
1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0
VIV
Component X and a cell of negligible internal resistance are placed in a circuit.
A variable resistor R is connected in series with component X. The ammeter reads 20mA.
4.0V
4.0V
Component X and the cell are now placed in a potential divider circuit.
(a) Outline why component X is considered non-ohmic. [1]
(b(i)) Determine the resistance of the variable resistor. [3]
(b(ii)) Calculate the power dissipated in the circuit. [1]
(c(i)) State the range of current that the ammeter can measure as the slider S of the potential divider
is moved from Q to P. [1]
(c(ii)) Describe, by reference to your answer for (c)(i), the advantage of the potential divider
arrangement over the arrangement in (b).
1.) Two long parallel current-carrying wires P and Q are separated by 0.10 m. The current in wire P is 5.0 A.
The magnetic force on a length of 0.50 m of wire P due to the current in wire Q is 2.0 × 10-s N.
(a) State and explain the magnitude of the force on a length of 0.50 m of wire Q due to the current in P. [2]
(b) Calculate the current in wire Q. [2]
(c) Another current-carrying wire R is placed parallel to wires P and Q and halfway between them as shown.
wire P
wire R
wire Q
0.05 m
0.05 m
The net magnetic force on wire Q is now zero.
(c.i) State the direction of the current in R, relative to the current in P.[1]
(c.ii) Deduce the current in R. [2]
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