Prelim Physics Thread (1 Viewer)

[eyeseeyou]

Use E = F/q.

Oh right right, is it always neccessary to include direction?

 

[jathu123]

Oh right right, is it always neccessary to include direction?

I guess so. Electric field strength is a vector, and so you do need direction

 

[jathu123]

The captain of a fishing boat uses an echo sounder to determine the depth of a school of fish below the boat. The captain finds that the reflected waves return after o.15s and o.20s. The captain believes the first reflectin to be from the fish while the seconds is from the ocean floor. If the speed of sound in seawater is 1440m/s, determine the depth of the sea floor. (3 marks)

can anyone tell me the expected way of doing a question like this for year 11 physics? i looked it up and found ways which correlate to actual echo sounders, but realise i've never done this stuff in class.

any help is greatly appreciated!

Drawing a small diagram may help. Since they want the depth of the sea floor, ignore the first reflection. So the waves return after 0.20s. This means that in 0.20s, the wave traveled from the surface to the sea floor, and back again to the surface. So if I let x be the depth of the sea floor, the wave basically travelled 2x in 0.20s. Now you the formula s=d/t to work out x.

1440 = (2x)/0.2
x = 144m

(if you have the answers, do make sure this is right)

 

[fjitlid]

thanks dude

 

[Jonothancroller]

Hey, I've done a question two ways and I have no idea which one is right, can someone explain why one is wrong.

A spaceship in deep space fires its rocket motor that exerts a force of 8.00 x 10^4 N. The rocket continues to fire while the spaceship moves through a distance of 1.0 x 10^7 m. If the 6.0 x 104 kg spaceship had been moving at 4.5 x 10^3 m s^-1 before the rocket motor was fired, how fast is it moving now?
Method 1:
Change in kinetic energy = Fs = 8x10^11 J
Therefore, (Δv)^2=(8x10^11)/(m/2) = 2.67x10^7, Δv = 5163 ms^-1
Therefore, final v=9663ms^-1
Method 2:
Total kinetic energy = 8x10^11+3x10^4x(4.5x10^3)^2 = 1.41x10^12 J
v = (1.41x10^12/(3x10^4))^(1/2) = 6855 ms^-1

 

[eyeseeyou]

Hey, I've done a question two ways and I have no idea which one is right, can someone explain why one is wrong.

A spaceship in deep space fires its rocket motor that exerts a force of 8.00 x 10^4 N. The rocket continues to fire while the spaceship moves through a distance of 1.0 x 10^7 m. If the 6.0 x 104 kg spaceship had been moving at 4.5 x 10^3 m s^-1 before the rocket motor was fired, how fast is it moving now?
Method 1:
Change in kinetic energy = Fs = 8x10^11 J
Therefore, (Δv)^2=(8x10^11)/(m/2) = 2.67x10^7, Δv = 5163 ms^-1
Therefore, final v=9663ms^-1
Method 2:
Total kinetic energy = 8x10^11+3x10^4x(4.5x10^3)^2 = 1.41x10^12 J
v = (1.41x10^12/(3x10^4))^(1/2) = 6855 ms^-1

in Method 1, you are assuming that the mass of the rocket is constant. It is not, it is losing fuel, and so its mass is actually decreasing. Therefore, you cannot say the change in kinetic energy is purely due to the change in velocity. Your second method is more appropriate and should be giving you the correct answer!

 

[calamebe]

in Method 1, you are assuming that the mass of the rocket is constant. It is not, it is losing fuel, and so its mass is actually decreasing. Therefore, you cannot say the change in kinetic energy is purely due to the change in velocity. Your second method is more appropriate and should be giving you the correct answer!

Well he would do that in method 2 as well anyway, as for the final kinetic, as using that Fs formula assumes that mass remains constant. To derive W = Fs, you must assume that mass is a constant, so that isn't why he got it wrong. To be able to accurately calculate it using the changing mass, you'd need to to find the rate at which it loses fuel (which might not even be constant), and know how much fuel it carries, etc. He just did a simple math error. It isn't true that (Δv)^2 = (8x10^11)/(m/2). Instead, (final velocity)^2 - (initial velocity)^2 = (8x10^11)/(m/2).

 

[Green Yoda]

A light globe that is just visible from a distance of 1.00 km can be observed using a telescope when it is placed on the Moon, a distance of 380 000 km. How many times brighter does the globe appear to be when it is 1.00 km away compared to when it is on the Moon?

 

[jathu123]

A light globe that is just visible from a distance of 1.00 km can be observed using a telescope when it is placed on the Moon, a distance of 380 000 km. How many times brighter does the globe appear to be when it is 1.00 km away compared to when it is on the Moon?

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