Answers by visitors named: Manish

First of all, Henry cannot be right because the ball was "thrown" at the ground. His formula is only valid if it was free fall. However, the ball gained acceleration from the throw. You have to use energy formulas. I have a similar question on my worksheet, but I don't have a mass so I think the my problem is wrong. You should use energy problems in your case, but I'm too lazy to do all the work. Here are the formulas you need. E=mgh E=(1/2)mv^2 Also, remember conservation of energy.

20km

15-8√3

Tan75/2

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unless you have a typo and meant: f(x) = 10,125(1.83)^x the second is exponential and can be written as g(r) = 9,638(1.95)^r clearly g(r) has a higher percentage change since 1.95 > 1.83 a more interesting question might have been: When will the amount be the same? 10,125(1.83)^x = 9,638(1.95)^x 10,125(1.83)^x = 9,638(1.95)^x (1.83)^x = .951901...(1.95)^x take log of both sides and use log rules xlog1.83 = log .951901... + xlog 1.95 x(log 1.83 - log 1.95) = log .951901... x = .776.. check: f(.776..) = 10,125(1.83)^.776.. = 1618410 g(.776..) = 9,638(1.95)^.776... = 16184.10

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Manish shrestha

The diameter of the head of a screw nails is 4cm and it's pitch 0.5cm. If the nails can magnify the applied effort 20 times, calculate it's MA, VR and efficiency