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...
Homework Statement
So in my school, the three AP science concentrations (Bio, Chem and Physics) are teaming together in groups to answer a question about water and it's relationship to life. So our team is doing this question: By changing the ratio between distilled water and the pollutant...
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!
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.