DatAtarLyfe
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M9, you never didI know. I may include a diagram of the apparatus later and attempt a rewording.
M9, you never didI know. I may include a diagram of the apparatus later and attempt a rewording.
Still trying to understand what these mean lolA student sets up the following apparatus.
Two wires are connected to a power box (6V) on one end, and a strip of aluminium foil (resistance = 1Ω) on the other. The aluminium fold bends on top of itself in this circuit. (Picture a C shape the foil is making.)
When the switch was turned on, momentarily the aluminium foil appeared to expand.
a) What happened to the power box once the switch got flicked and why? (2)
b) Carefully explain (with or without use of a diagram) why the aluminium foil expanded. (4)
c) The average seperation distance of the foil was 3mm. Calculate the force each side of the foil exerted on each other. (3)
I got lazy. I'll do one today lol.M9, you never did
Well can you post another qu. we can do whilst you work that out?I got lazy. I'll do one today lol.
a) I'm a bit confused to as what you mean lol but I'll try. As the current from the power source does not automatically go to maximum and requires a momentary build-up time, the reading on the voltmeter (power box?) will similarly increase as V is proportional to I (shown in V = IR, where R is a constant in this case). That is, as I increases to a maximum, V also increases to a maximum.A student sets up the following apparatus.
Two wires are connected to a power box (6V) on one end, and a strip of aluminium foil (resistance = 1Ω) on the other. The aluminium fold bends on top of itself in this circuit. (Picture a C shape the foil is making.)
When the switch was turned on, momentarily the aluminium foil appeared to expand.
a) What happened to the power box once the switch got flicked and why? (2)
b) Carefully explain (with or without use of a diagram) why the aluminium foil expanded. (4)
c) The average seperation distance of the foil was 3mm. Calculate the force each side of the foil exerted on each other. (3)
My bad on c), was meant to say force per metre. That aside since you didn't directly show your substitution given what you had I will only assume you're right. Also, units, Newtons.a) I'm a bit confused to as what you mean lol but I'll try. As the current from the power source does not automatically go to maximum and requires a momentary build-up time, the reading on the voltmeter (power box?) will similarly increase as V is proportional to I (shown in V = IR, where R is a constant in this case). That is, as I increases to a maximum, V also increases to a maximum.
b) Parallel current carrying conductors produce a force which depending on the direction of each current will either repel the two conductors from each other or attract them. This relationship is shown in the equation F/l = k.I1.I2/d. In this case, as the currents are in opposite directions, a force of repulsion will act on them, causing the loop of aluminium foil to 'expand' as the sides are forces away from each other. This force is produced as a result of the motor effect, where the interaction of each side of the loops magnetic field interacts with the other, causing a force to be produced. This can be modeled with the right hand grip and right hand palm rules demonstrating the repulsive force. The expansion is only momentary as according to the formula F/l = k.I1.I2/d, the force is inversely proportional to distance, and as the coil expands, the system reaches equilibrium as the force becomes negligible.
c) idk how to calculate this without the length of at least one side of the aluminium. But V = IR hence I = 6A, therefore F = l.k.I1.I2/d
F = 2.4 x 10^(-3) x length away from each other side of the aluminium loop
yeah i would set it out properly if i was using latex or writing it lol but i ceebsedMy bad on c), was meant to say force per metre. That aside since you didn't directly show your substitution given what you had I will only assume you're right. Also, units, Newtons.
ah crap, thats so obvious lol. aluminium would melt yeaha) - That was testing your skills. That is a short circuit right there so the overload on the powerbox will, in a matter of milliseconds, be flicked off.
tyb) was well done
lol hsc questionNEXT QUESTION:
a) Outline Galileo's analysis of projectile motion. (3 mks)
b) An astronaut on the Moon throws a stone from a 150m high cliff. The stone hits the ground a distance of 300m away, 21.0s later. If gravitational acceleration is equal to 1.6ms^-2 on the Moon, determine the initial velocity which the stone was projected from. (5 mks)
c) Suppose the exact same scenario was replicated, except on Earth. Identify two differences between the motion of the stone on both planets. (2 mks)
I tweaked b) a bit.lol hsc question
a) Galileo was the first person to mathematically analyse projectile motion and separate it into xNEXT QUESTION:
a) Outline Galileo's analysis of projectile motion. (3 mks)
b) An astronaut on the Moon throws a stone from a 150m high cliff. The stone hits the ground a distance of 300m away, 21.0s later. If gravitational acceleration is equal to 1.6ms^-2 on the Moon, determine the initial velocity which the stone was projected from. (5 mks)
c) Suppose the exact same scenario was replicated, except on Earth. Identify two differences between the motion of the stone on both planets. (2 mks)
Lol just use daum like I do.a) Galileo was the first person to mathematically analyse projectile motion and separate it into x
and y components. The main problem with previous analysis of the motion had to do with the
objects moving too fast for accurate measurement and the presence of air resistance. However through
experimentation with ball-bearings and inclined planes Galileo was able to conclude that projectile
motion was made up of two separate components – a constant horizontal motion with zero
acceleration and an accelerating vertical motion with a uniform acceleration (gravity). He also
realized that motion of projectiles has a parabolic trajectory due to the proportionality of the
constant x component and the uniform acceleration of the y component.
b)
c) The maximum height would be lower, The range would be lower
spend like 3x more time writing up the tex than i did doing all three questions combined
I do use daum lol, thats why it takes so longLol just use daum like I do.
For a) you should always mention that this is true if and only if air resistance can be ignored.
b) Should've given answer to 2 sig fig; least amount of sig figs in question
Other than that beautiful
I'd agree with this - you might as well get into the habit of writing the correct number of significant figures.True, but unless you're InteGrand or someone it's faster than actually typing the LaTeX
Depends on who marks your question. Some examiners get fussy others don't. That's why it depends on a chance you can't risk