Question

Show all working clearly and include diagrams where appropriate. Express all answers to the correct number of significant figures and units. A storm approaches an office and the air pressure outside suddenly drops from 1.013 kPa to 0.960 kPa. Before the pressure inside the office has time to change, what is the magnitude of the overall force exerted on an external window of the office with dimensions 1.2 m by 1.8 m? Four people, with a combined mass of 320 kg, wish to build a raft that will carry them over a river. There is a plentiful supply of cylindrical timber logs available, each one with a radius of 0.08 m and a length of 3.00 m. Given that the density of the timber in the logs is 725 kg/m3, calculate the least number of logs needed to form this raft? What is the minimum amount of heat that must be put into a 3.2 kg block of ice at -12 degree C to convert it to water with a temperature of 5 degree C? After my car has been parked outside on a cold night, the metal panels will feel colder than those made from plastic. Explain why this is so. How much energy is needed to move a charge of 5e through a potential difference of 20 V? A charge of 6 is fixed 11 cm away from another fixed charge of -4 muC. If a third charge of -3 mu Cwas placed on the line between the first two charges so that it was 5 cm from the first and 6 cm from the second (as shown below), what force (magnitude and direction) would the third charge experience? 6mu C -3mu C -4 mu C _ 5 cm__ 6 cm _ Place a charged comb near a thin stream of running water from a tap. (for information regarding how to charge the comb see section 4.1.1 of your study book). Draw a labelled diagram to illustrate your observation. Describe what you observed. Give an explanation of your observations. (Hint: Consider whether or not the stream of water is charged.) You may need to do a little research for this answer. Revision questions: A ball is tossed straight up into the air and reaches a maximum height of 15 m. Ignoring the effect of air resistance calculate the following: How long after the ball is thrown should it take before it returns? How fast will the ball be travelling just before it lands?

Answer 1

1) F = $\Delta$P A =

= (1.013-0.960)x(1.2x1.3) = 0.11448 kN

2) bouyant force > W_M + W_L

=> $\rho$_water x(n x pi R²L)g > 320 xg + $\rho$_log x(n x pi R²L)g

=> 16.579 n > 320

=> n_min = 19.3 = 20 logs

3) Q = Q_{(-12 to 0)} + Q_{(s to L)} +Q_{(0 to 5)} = (3.2 x 2.108 x 12) kJ + (3.2 x 334) kJ + (3.2 x 4.186 x 5) kJ = 1216.7 kJ

4) When heat is transferred from one material to another it is called conduction. Metal is a good conductor of heat, while nonmetals like wood and plastic are poor conductors. The temperature of any metal object in any environment is about the same as the air that surrounds it. But in case of non metals there is a temperature difference.

When you touch a metal object that's surrounded by air cooler than your body, the object quickly conducts heat away from your fingertips, which makes them feel cold. The sensation travels to your brain, which perceives that the metal object is cold. But in case of plastic they don't adapt to the surroundings because of poor conductivity.

5) E= W = q x V = 5e x 20 V = 100 eV or 100x 1.6x10^-19 J = 1.6x 10^-17 J

6) F = (Kq₁q₃/r₁² ) i + (Kq₂q₃/r₂² )i      where i is a unit vector in positive X direction that is toeard -4 $\mu$C

= (9x10⁹ x 6 x10^-6 x 3x10^-6/0.05² ) -i + (9x10⁹ x 4 x10^-6 x 3x10^-6/0.06² ) -i

= 94.8 N (-i)      or 94.8 N towards 6$\mu$C

7) a)

The stream of water deviates a little towards the comb.

b) Water has no net electrical charge, but the negatively charged comb induces a complementary charge in the water. The electrons in the water can move somewhat. When the negatively charged comb comes near the water, it repels the electrons, so that the side of the water nearest the comb then has a positive charge. The attraction between this positive charge and the negatively charged comb results in a net force on the water, bending the stream.

8) v² - u² = 2aS => 0 - u² = 2x-9.8 x 15

=> u = 17.146 m/s

therefore time to reach maximum height = (v-u) / a = (0-17.146)/-9.8 = 1.7496 s

a) so total time taken by ball to return = 2 x 1.7496 = 3.499 s

b) V before landing = velocity with which it was thrown = u = 17.146m/s