21ST CENT.ASTRONOMY(LL)W/CODE WKBK PKG.
6th Edition
ISBN: 9780393874921
Author: PALEN
Publisher: Norton, W. W. & Company, Inc.
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Chapter 9, Problem 17QP
To determine
How the secondary atmospheres of the terrestrial planets were created.
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Students have asked these similar questions
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]
Chapter 9 Solutions
21ST CENT.ASTRONOMY(LL)W/CODE WKBK PKG.
Ch. 9.1 - Prob. 9.1CYUCh. 9.2 - Prob. 9.2CYUCh. 9.3 - Prob. 9.3ACYUCh. 9.3 - Prob. 9.3BCYUCh. 9.4 - Prob. 9.4CYUCh. 9.5 - Prob. 9.5CYUCh. 9 - Prob. 1QPCh. 9 - Prob. 2QPCh. 9 - Prob. 3QPCh. 9 - Prob. 4QP
Ch. 9 - Prob. 5QPCh. 9 - Prob. 6QPCh. 9 - Prob. 7QPCh. 9 - Prob. 8QPCh. 9 - Prob. 9QPCh. 9 - Prob. 10QPCh. 9 - Prob. 11QPCh. 9 - Prob. 12QPCh. 9 - Prob. 13QPCh. 9 - Prob. 14QPCh. 9 - Prob. 15QPCh. 9 - Prob. 16QPCh. 9 - Prob. 17QPCh. 9 - Prob. 18QPCh. 9 - Prob. 19QPCh. 9 - Prob. 20QPCh. 9 - Prob. 21QPCh. 9 - Prob. 22QPCh. 9 - Prob. 23QPCh. 9 - Prob. 24QPCh. 9 - Prob. 25QPCh. 9 - Prob. 26QPCh. 9 - Prob. 27QPCh. 9 - Prob. 28QPCh. 9 - Prob. 29QPCh. 9 - Prob. 30QPCh. 9 - Prob. 31QPCh. 9 - Prob. 32QPCh. 9 - Prob. 33QPCh. 9 - Prob. 34QPCh. 9 - Prob. 35QPCh. 9 - Prob. 36QPCh. 9 - Prob. 37QPCh. 9 - Prob. 38QPCh. 9 - Prob. 39QPCh. 9 - Prob. 40QPCh. 9 - Prob. 41QPCh. 9 - Prob. 42QPCh. 9 - Prob. 43QPCh. 9 - Prob. 44QPCh. 9 - Prob. 45QP
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