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I was just wondering the principle of a hydrometer. i know its based on Archimedes principle/ the law of flotation but i can't seem to understand how this explains the fact that for a denser liquid, the object is buoyed up more and in a less dense liquid it would sink more. i hope someone can help me clarify this. Thanks.

(i) how many different poker hands can be dealt to a player? (ii) how many hands include the Ace of spades? (iii) how many hands do not contain include the ace of spades? (iv)how many hands include all 4 aces? (v) how many hands contain exactly 3 aces? so i know that there are 52 cards in a pack and that a poker hand is 5 cards, so the answer to (i) is 52 choose 5 52C5 , using the calculator i get 2598960. and i know that there is only one ace of spades in a pack of cards but i'm not quite sure how to work out part (ii) and if i knew the answer to this then obviously i could work out the answer to part (iii) could anyone give me a bit of help please? thanks.

A skateboarder with a total mass of 70 kg starts from rest at the top of a ramp and accelerates down it. The ramp is 25 m long and is at an angle of 200 to the horizontal. The skateboarder has a velocity of 12.2 m s–1 at the bottom of the ramp. (i) Calculate the average acceleration of the skateboarder on the ramp. (ii) Calculate the component of the skateboarder’s weight that is parallel to the ramp. using one of the equations of motion i can get the acceleration to be 2.98 ms-2 but i,m having trouble with the second part , i know that the vertical component (his weight due to gravity) is (70)(9.8) = 686N. can i use trigonometry or something to get the the component parallel to the ramp?

well if i factorise by completing the square then i get x2 + 2gx + g2 =0 which gives me (x+g)(x+g) = 0 like as you said (x+a)(x+b) = 0 and g + g =2g (a + b = 2g as you said ) and ab = (g)(g) = g2 =c , but why is it equal to c?

if x2+y2+2gx + 2fy +c =0 is on the x-axis, prove that g2 = c. I know that on the x-axis, y=0. x2 + (0)2 +2gx + 2f(0) +c =0 x2 + 2gx +c= 0 (which is a quadratic) but now i'm stuck. solving the quadratic with the quadratic equation formula doesn't help, so what should i do?

i think i get where your coming from, the bike has already just passed the point when the car just takes off so for a short space of time, the bike is in front of the car, until the car accelerates enough to catch up, thanks for your help.

well yes I did realise that thats what they were asking, but i cant seem to visualise the situation, i drew it out but it doesnt seem right to me, if the rear of the bike and the front of the car are at P at the same instant how could it catch up if they're already at the same point P? sorry note taken , but there are my ideas so far, still need clarification, thanks .

I'm having a little trouble with this problem for my physics homework, could anyone give me a step in the right direction here, thanks. ''The rear of a bicycle passes a point P on a road and travels at a steady speed of 12 m/s down the road. At the same instant the front of a car starts from rest at P and moves in the same direction as the bicycle with an acceleration of 2 m/s^2. When and how far from P does the front of the car catch up with the rear of the bicycle.''