Recent content by ericcy

And my apologies, the equation pasted in wrong, they did t=5[distance to island]/(40[velocity relative to water]-6[velocity of water relative to shore]=0.15h)

If we were to solve for time, like in this question, would we always be expected to use the velocity of the boat relative to the shore and not the water?

I've looked it up online and someone did t=40−65=0.15(h) I was just wondering why they would subtract the velocities. Could something explain this to me please? thanks.

That's the data given in the question, so I dunno...

d1x= sin58(618)=-524.093km, d1y=cos58(618)=327.49km d2x= cos35(361)= 295.713km, d2y= sin35(361)=-207.061km drx= -524.093+295.713= -228.38km dry= 327.49-207.061= 120.429km Using pythagoras with these values I get dr= 258km and an angle (using tan) N62W @PeroK

What do you mean a sign convention? Also, is everything else alright? I can clarify some numbers for you if you need it.

I apologize if my work is messy or hard to read. Reply above explains what it means, thanks.

Broke it into its components finding d1x, d1y, d2x, etc... Using those components I found drx to be 228.38km and dry to be 120.429km. Did Pythagoras to get 258km as the resultant displacement, heading N62W. I'm honestly lost. I'm doing the question the correct way, I just don't know what I'm...

I can't think of it. Maybe d=1/2at^2?

All I can think of is that kinematics equation from earlier. We don't have velocity so that's the only way I can think of.

The down-slope force acting on the object would be fgx, so mgSin(theta)= 1.226m Fgy is mgCos(theta)= 9.733m Fn=Fgy so Fn= 9.733m Ff=Fn(mu) Ff=9.733m(0.11)= 1.0706 Ff=1.0706m So the down-slope force would be 1.226m while the upward force is 1.0706m Not sure how this can help me solve for time...

Could I do d=Vit+1/2at^2? I could use the length from the dimensions, 12m. Would that be the proper way to tackle this question?

Turns out for this one I was just looking in the wrong section, lol. But there are quite a few mistakes in the back of the book that I've noticed. That's kind of why I've been asking a lot of questions on here to make sure that I'm right and the book is wrong. The title of the book is Physics...

Because the friction is the same in both parts, the calculated acceleration from (b) should be the same for (c) I knew I could find Vf, and thought I could do it with an energy equation Ei=Ef mgh=1/2mv^2 gh=1/2v^2 (2)9.81(1.5)=1/2v^2(2) (square root)29.43=(square root)v^2 v= 5.424 Then...

I determined 42m to be the hypotenuse so I used sine law to find the height of the incline, 10.87m. I used this height in the equation Ei=Ef, since they should be equal. Ei=Ef mgh=1/2mv^2 (at the start there is no kinetic energy, at rest. at the end there is only kinetic, no potential)...