Recent content by zaga04

Homework Statement Heat input of an engine is 1330 J/s at 375°C and heat output is 800 J/s. Each second, how much work is done in the process?Homework Equations QL/QH=1-eThe Attempt at a Solution QL=(1-e)QH QH=W/e W=average of 1330 and 800 which is 1065J QH=1065J/.40 (e was calculated in...

Homework Statement 1. A 5kg object has a momentum of 15 kg m/s. What is the net force required to accelerate the object to 8 m/s over 15 seconds? A) .35N B) 1.7N C) 5.1 N D) 11 N E) 23 N Homework Equations F=deltaP/deltaT The Attempt at a Solution So I use that formula (which...

Multiple Choice Question: What are the units of the spring constant K? I. N/m^2 II. kg/s^2 III. J/m^2 Choices: A. I only B. II only C. III only D. I and II E. II and III I cannot find it in my textbook for some reason...

So ignoring Fp would it go something like this... Box 1: Ft-mgsin30=m1a1 Ft=m1a1 + mgsin30 Box 2: Ft-mgsin60=m2a2 m1a1+mgsin30-mgsin60 = m2a2 mgsin30-mgsin60/m1+m2 = a a= -1.8 m/s^2 ------------------------- b. Box 1 Ft-mgsin30-ukFn=ma Ft=ma + mgsin30+ukFn *Fn=mgcos30=16.97* Box 2...

Alright, I'll go see if I can solve it without having an applied force.

Homework Statement They're two boxes connected on a pulley, each weighing 2 kg. Box 1 is on a 30 degree incline and box 2 is on a 60 degree incline. a. Find the acceleration ignoring friction b. Find the acceleration if the coefficient of kinetic friction is 0.1. Homework Equations F=ma...

So would it just be 8+10-15-18 since they are at 90? They wouldn't have any x and y components since they are not at an angle. Also what would the resultant angle be?

Homework Statement An object is at the origin. They're four perpendicular forces acting upon it. In the negative x axis, 18 N. positive x axis, 10 N. negative y axis, 8 N. and positive y-axis 15 N. calculate the resultant force. What angle is it at? Homework Equations F=Ma? The...

Ok that makes sense. So is this how you do it? BOX 2: ∑Fy=Fnb-m2gcos38-m2ay=0 Fnb=m2gcos38=10.04 N ∑Fx=T-Fb-m2gsin38=m2ax (Fb=ukfn) T=m2ax + Fb + m2gsin38 m2ax + ubFnb+m2gsin38 m2ax + (.45)(10.04)+(1.3)(9.8)sin38 T=m2ax + 12.36 N BOX 1: ∑Fy=Fna-m1gcos38-Fnb = m1ay=0 Fna=m1gcos38 + Fnb...

Is this how you would do it? so M=box1, m=box 2 a=Mgsin38-.35Mg38-t/M a=mgsin38-.45mgcos38-t/m solve for T, and get 2.06 N then plug it back in the equation and get 4.8 m/s^2? can you please tell me if this is right or not. or can you please help me get started if it's wrong!

OK thanks, but what is the equation for tension?

Do you have to find the tension from the string to solve it? I am not sure how to find the tension if so. I am still a little confused on how to solve the problem.

Homework Statement They're two boxes on a ramp. Box 1 (on the bottom) is 8.60kg and has a coefficient of kinetic friction of .35 with an incline plane. Box 2 (on top of Box 1) is 1.3kg and has a coefficient of kinetic friction of .45 between the boxes. The two boxes are connected through a...

I put you can drop the ball and use the timer to measure the amount of time it took to hit the bottom. Then you know acceleration (9.8 m/s^2) so you can use some formula which I cannot think of right to calculate the depth.

This is not a homework problem, I am not sure what section to put this (I apologize if it's in the wrong place). So I take AP Physics B at my high school and we just got our first test back on Kinematics in One Dimension and I got a 53%! I know that might sound horrific (which it is) but my...