high school physics homework help

A:I don't know how to do it except using trig. You know the hypotenuse and an angle. So SIN 37 = Opposite/Hypotenuse = y/10. Thus y = 10*SIN 37 = 6.01815. That is your vertical component. COS 37 = Adjacent/Hypotenuse = x/10. x = 10*COS 37 = 7.986. That is your horizontal component.

A:At the top of its motion it speed should be O and the acceleration should be 0. It has hit the point of where it is about to accelerate downward when it starts to descend. Sort of like if a person climbed to the top of a hill - when they get to the top they will stop temporarily and then descend - at the top the speed will be zero as well as the acceleration. As the start the movement back the acceleration and speed will increase.

A:Other answer is good. Forces can be reduced to their components along the coordinate axes, then added. The final two components make up the two sides (for 2 dimensions) of the triangle. So 1) reduce all the forces to their x & y components (best to draw a rough diagram first to make sure you're getting the signs right) 2) add the x's together 3) add the y's together the final force is in the direction of the angle A where X/Y=tan(A) (or is it Y/X .. i've forgotten!!!) and its magnitude is Sqrt(X^2 +Y^2) - thats the hypotenuse of the triangle, don't you know. Tricky part is to keep all the signs of the components correct. Pick your origin and x&y axes and if (in your rough drawing) the force is below the axis the y component better be negative, if the force is to the left (of the origin) then the x comp. is negative, right? Sin(a)=opp/hyp, Cos(a)=adj/hyp Must use calculator to figure arctan(X/Y)=A (or tangent tables) You'll have to find out if tan is opp/adj (X/Y) or vice versa, sorry. To the second part. I guess you know the formulas for distance and velocity which includes acceleration? d=d0+v0*t+1/2*a*t^2 and v=v0+a*t, right? ALL you have to do is know that both of these work with components, too. So since you already know the force ( X & Y) and since F=ma you can do all four equations by adding the components. dy (distance in y direction) = dy0 +vy * t + 1/2*a*t^2 so you know t , you figured ay=Fy/m and you can figure the y component of the given velocity. Figure the vx and ax the same way. Solve for dy, dx, vy,vx and do the two triangels for direction and magnitude (the magnitude of v is the speed, of the d is the distance) like above.

A:If your daughter has trouble in a subject, your help is definitely warranted. I would tell her that you want the best for her and that it is important that she understand physics. Give her an example of how you use physics in everyday life. Cooking is always a great example. As far as the SAT is concerned, she has to take a practice test and see where she is. Nagging her to study for a comprehensive test is really not useful and yes I would consider it nagging too. After she takes the practice test and finds her weak spots, she can study in those areas and make sure she is ready to take the real test. Take care, Troy

A:MIND!! PEOPLE WOULD LIVE INSIDE THE WHEEL TUBE, THEIR FEET ON THE OUTER DIAMETER, THEIR HEADS POINTED TO THE WHEEL AXIS! Centripetal acceleration g=R*w^2, R=1600/2=800 m, angular speed to be found when g=9.8 m/s^2 exactly, hence w=sqrt(g/R)=sqrt(9.8/800)=0,1107 rad/s. To reach the angular speed w in t=2h=7200s, we must apply angular acceleration w’=w/t=0.1107/7200=1.537*10^-5 rad/s^2. Inertia moment of the wheel is I=m*Rm^2, m=187016kg, mean radius Rm=(1600-1500)/2 m, so I=4.68*10^8 kg*m^2. Thus torque T=I*w’=7186 Nm – never believe us! Check it!