1(b) If the electric bulb has 51% effiency and the energy supplied by the dry cell is 90 J/min, what is the useful power output of the bulb?
2(a) A diver jumps from the springboard at the swimming pool and dives into the water below.
(i) Assuming that the maximum height that the diver reaches is 5.5m above the ground, calculate his speed when he hits the water.
(ii) Calculate the height of the springboard above the ground, given that the lowest speed that the diver can reach as he touches the water is 10 m/s.
Please label your questions clearly and explain how you get the steps and answers as I am very bad at this topic. The person who can give me the most clearest and understandable answer will get ten points. Thank you for your cooperation.Physics Ultimate Challenge?1) Chemical potential will be converted into electrical energy. This energy will then be convered to 2 forms of energy. One is light which is useful energy and the other is heat. Heat is given out by the filament and is a waste.
b) 51% * 90= 4.9J/min
2)
i) They have only given u the mass of the guy and want u to find out the speed at which he hits the water
When the guy reaches maximum height then he has maximum potential energy(P.E) The formula is mgh. Now when he falls almost all the PE energy is converted to Kinetic Energy (KE). The formula is (0.5*m*v*v) Some of the energy is lost has be converted to heat and sound enery when he hits the water but you can exclude that.
So as sid before KE=PE. The m's can be canceled since they are equal. This leaves us with gh=0.5*v*v. To find the speed rearrnage the equation to give 2gh=v^2. this can be furthered reduced to v= suare root (2gh). Now replace g with 9.81 and h with 5.5 . Mulitply the 2 values and suare root it to give 10.38 m/s.
sorry but i am not sure how to do the alst question.Physics Ultimate Challenge?1. Chemical potential -%26gt; electrical -%26gt; light + thermal
note that efficiency is defined as work output / work input.
work input = 90
work output = 90 * 51%
2. The maximum height occurs when the velocity in vertical direction is 0.
0 = v0 + at
t = v0/g where a = -g
s = s0 + v0t + at^2 /2
= v0(v0/g) - g(v0/g)^2 / 2 = 5.5
= v0^2/g - v0^2/(2g) = 5.5
=v0^2/(2g)
v0 = sqrt[11g]
v = v0 + at
= sqrt[11g] - gt
where t satisfies the equation
0 = sqrt(11g)t - (g/2)t^2 = t[(sqrt(11g) - gt/2]; t = 2sqrt(11g)/g
So
v = sqrt[11g] - 2 sqrt[11g] = -sqrt(11g)
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Note that an easier method is to use the conservation of energy with PE = KE as the answerer below noted.
So mgh = mg*5.5 = mv^2/2 or 5.5g = v^2/2; v = - sqrt[11g] in agreement with the answer attained by the above method. Note that v must be negative as downward is the negative direction.
