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alright so the velocity is 12 km/h and the distance is 1.5 km 1.5 km/ 12 km/h = 0.125 h

Homework Statement Popeye was sailing his boat in a straight channel one half km wide. The steady current flows south at four km/h and a wind blows from the north, parallel to the current. Popeye starts from the west shore. He sets his sail and tiller so that he moves diagonally across and up...

Well i was thinking of the orientation shouldn't it be like -6.4% see: Profit % = mp(adown(down)+aup(up))/p(mg(down)-ma(down)) x 100% Profit % = mp(adown+aup)/mp (g-a) x 100% Profit % = (adown+aup)/(g-a) x 100% Profit % = (2 m/s^2(down) - 2.5 m/s^2(down)) / (9.8m/s^2 (down)- 2 m/s^2(down))...

I would like to thank you for all your help. You have been extremely patient and once again thank you.

Profit % = mp(adown+aup)/p(mg-ma) x 100% Profit % = mp(adown+aup)/mp (g-a) x 100% Profit % = (adown+aup)/(g-a) x 100% Profit % = (2 m/s^2 + 2.5 m/s^2) / (9.8m/s^2 - 2 m/s^2) x 100% Profit % = 4.5 m/s^2 / 7.8 m/s^2 x 100% Profit % = 57.7 %

Profit % = mp(adown+aup)/p(mg-ma) x 100%

if u mean just the price then Profit % = mp(adown+aup)/mp x 100% and for what they paid in acceleration wise Profit % = mp(adown+aup)/mp(g+aup) x 100%

okay then Profit % = mp(adown+aup)/mp x 100%, is this correct?

umm, i understand what you wrote but like if u want me to be hundred percent accurate i cannot be because i don't know how much the price is or the mass.

well interms of actual numbers.. they make a profit of 4.5 but like in percentage i am not sure how to find out.

Wouldn't it be the same thing except for the sign so like this p(mg-ma)

p(mg+ ma) he will pay whatever the weight the balance will show. since the balance shows the weight as mg+ ma as accelerating, and he buys it, to the weight then it would be p(mg+ ma) right?

So if they are selling the price for mgp than they are actually just selling it for the acceleration they go up by mass

Fnet + Fn = Fg ma + mg = mg

The spring balance must be exerting the normal force