The new force and the new power input

Question

Want to see the full answer?

Answer 1

Question

Chapter 23.2, Problem 23.4QQ

To determine

The new force and the new power input

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

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]

2.) A 50.0 resistor is connected to a cell of emf 3.00 V. The voltmeter and the ammeter in the circuit are ideal. V A 50.00 (a) The current in the ammeter is 59.0 mA. Calculate the internal resistance of the cell. The circuit is changed by connecting another resistor R in parallel to the 50.0 resistor. V A 50.00 R (b) Explain the effect of this change on R is made of a resistive wire of uniform cross-sectional area 3.1 × 10-8 m², resistivity 4.9 × 10-70m and length L. The resistance of R is given by the equation R = KL where k is a constant. (b.i) the reading of the ammeter. [2] (b.ii) the reading of the voltmeter. [2] (c) Calculate k. State an appropriate unit for your answer. [3] [2]