F4 phy

1. Energy = Force * Displacement
(J) = (N) * (m)
Force = Mass * Acceleration
(N) = (kg) * (m/s²)

So, (J) = (kg*m/s²)*m
= kg m² / s²
So the answer is C.

2. Use v² - u² = 2as
s = 0 since it returns to the same position.
v = -u.
That means the velocity of the ball when returning to the starting point is equal but opposite to the initial velocity.
Therefore, change of velocity = v - (-v) = 2v.

Throughout the fall, the acceleration is the same as the gravity g. So the change of acceleration is (g - g) = 0

So the answer is C.

3. If melting ice is NOT used, extra energy is needed to bring the ice from the temperature below the freezing point from the refrigerator to the melting point. So the energy measured will be larger than the energy actually used as latent heat to melt the ice. Therefore, melting ice should be used in order to measure the specific latent heat accurately.



2007-01-08 21:18:25 補充：
Do you still not understand the explaination about Q.3?Let Ti be the temp. of the ice just taken from the fridge.Energy measured in melting m kg of ice= m*c*(0-Ti) + m*l. Note that c is the specific heat capacity for ice, which is NOT 4200 J/kg/C.

2007-01-08 21:18:59 補充：
If melting ice is used, Ti = 0. So the energy consumed = m*l. So latent heat l can be obtained accurately by (Energy / m).Got it?

1. A joule is the work done or energy required to exert a force of one newton for a distance of one metre (http://en.wikipedia.org/wiki/Joule)
so 1J =1Nm
by converting F=ma to its units... 1N=1kgms^-2
combining the two... J=kgms^-2 x m = kgm^2s^-2
ie C

2. by v^2=u^2 + 2as.... s=0 (returns to starting point)
so v^2=u^2
since the direction of u and v must be different (u is upwards n v is downwards), we know that u = -v
change in velocity = v-u = v- (-v) = v+v = 2v
acceleration always = g, so change in acceleration = 0
again, C

3. because ice immediately taken out of a refrigerator can have a temperature lower than 0C, if we use this ice, we would be measuring the amount of energy needed to bring the ice to 0C (specific heat capacity of ice) together with the energy required to melt the ice (specific latent heat of fusion)

3 is a very common question and i remember there should be a model answer to that, you may look it up in the past paper since i took CE in 03.... =.=