Homework Statement
Two identical loudspeakers
are driven in phase by the same amplifier at a frequency of 680 Hz. The
speakers are 4.6 m apart. An observer stands 9 m away
from one of the speakers as shown. The observer
then starts moving directly towards the closest speaker.
How far does the...
Homework Statement
A molecule of DNA (deoxyribonucleic acid) is 2.10 μm long. The ends of the molecule become singly ionized: negative on one end, positive on the other. The helical molecule acts like a spring and compresses 1.09% upon becoming charged. Determine the effective spring constant...
Homework Statement
For the reaction below, the constant pressure heat of reaction is qp = −3256 kJ mol−1 at 25 °C. What is the constant volume heat of reaction, qV , at 25 °C?
16 CO(g) + 33 H2(g) ⟶ C16H34(l) + 16 H2O(l)
Enter your answer in kJ mol−1, rounded to the nearest kilojoule...
Homework Statement
A gas sample containing 3.00 moles of Helium gas undergoes a state change from 30 degrees celsius and 25.0 L to 45 degrees celsius and 15.0 L. What is the entropy change for the gas (ideal gas)? For He, Cp = 20.8 J/k*mol
Homework Equations
ΔS = Cv*ln(Tf/Ti) + nR*ln(Vf/Vi) =...
It is equal to 1 atm * 19.367 L
So to convert 19.367 atm*L to joules, I will multiply by 101.325 J
so work is equal to +1962.36 J ?
Edit: so to get the change in internal energy I just simply add work to q
1. Homework Statement
What are the values of q, w, ΔU, ΔH for the following constant pressure process for a system containing 0.596 moles of CH3OH ?
CH3OH(g, 123.0 ºC, 1.00 atm) ⟶ CH3OH(l, 30.0 ºC, 1.00 atm)
Molar heat capacity for CH3OH(g), Cp,m = 44.1 J K−1 mol−1
Molar heat capacity...
Homework Statement
If I am asked to give my answer in acceptable SI units and to 3 significant figures, how would I express my answer?
Homework Equations
Answer: 589883.4263 J
The Attempt at a Solution
My instinct would be to put this in KJ, but I don't know if that's an "acceptable SI...
Homework Statement
A flower pot is knocked off a window ledge from a height
d = 21.6 m
above the sidewalk as shown in the figure below. It falls toward an unsuspecting man of height
h = 1.71 m
who is standing below. Assume the man below requires 0.300 s to respond to a warning. How close to...
Homework Statement
What is w when a gas is compressed from 42.1 L to 25.1 L using a constant external pressure of 739 Torr? Remember to include a "+" or "−" sign as appropriate.
Homework Equations
W = -P(dV)
The Attempt at a Solution
Pressure = (739 Torr / 760 Torr)*(101.325 kPa)
W =...
Ah, thanks, I finally got the correct answer, but what do you mean I didn't have to find the oscillation frequency?
I feel a bit stupid to say this but the equations in my textbook are of the form: a = -Aω^2cos(2pi*f*t)
Homework Statement
A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 6.30 N is applied. A 0.540-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x = 5.00 cm and...
Oh, I see, so:
Fi = 9.81*2000 = 19600 N
Ff = 9.81*2325 = 3208.25 N
The change in force is 3208.25 N.
So 3208 = kx
k = 3208/0.0055
k = 583272.7273 N/m
Is this correct? If so, then I would just have to plug the numbers in for part B?
Thanks a lot
Homework Statement
A child's toy consists of a piece of plastic attached to a spring in the figure below. The spring is compressed against the floor a distance of 2.00 cm, and the toy is released. If the toy has a mass of 117 g and rises to a maximum height of 45.0 cm, estimate the force...
Homework Statement
When four people with a combined mass of 325 kg sit down in a 2000-kg car, they find that their weight compresses the springs an additional 0.55 cm.
A) What is the effective force constant of the springs?
B) The four people get out of the car and bounce it up and down. What...
1/2 * (mass of child + mass of merry-go-round) * (r*omega)2 - 1/2mv2 = 0
Does this look better? I'm still not getting the right answer but it's much closer...maybe I messed up the calculation or the answer is wrong...but is this correct?
Homework Statement
I have a basic problem where a child jumps tangentially onto the outer edge of a stationary merry-go-round, and you have to use conservation of momentum to find the final angular speed of the merry-go-round.
But the next part of the question asks "how much mechanical energy...
I think the FHx changes, and also, I think the torque from the weight of the beam changes, but is it still mg, or is it mg multiplied by an angle? I'm just having a really hard time making sense of what changes and what doesn't.
Homework Statement
"A beam of mass M = 280kg and length L = 2.2m is attached to a wall with a hinge (at a 90 degree angle to the wall, sticking out horizontally), and is supported at the other end by a wire making an angle of 30 degrees with the horizontal beam.
What is the force acting on...
Homework Statement
A 25kg child is spinning on a merry-go-round of mass 150kg and radius 2m at a constant angular velocity of 1rev/s. The child slowly walks to the center of the merry-go-round. Treat the child as a point mass and the merry-go-round as a uniform solid disk, and neglect friction...
Well KE at the top of the ramp is zero, but KE initial (initial being the start of the horizontal plane) must be equal to mgh, because all the mgh was converted into KE...
The ramp is frictionless, so there is no work done there, but the work done by gravity is mgh, right?
So just looking at...
Homework Statement
1) A 50kg person drives a car at 8.3m/s over a hump in the road. At the top of the hump, the driver feels a force of 143 N from the seat. What is the radius of the hump?
2) At what speed will the car need to move over the hump for the person to feel weightless at the top...
I can say that energy at the top of the ramp is equal to mgh, and when it gets to the bottom of the ramp, it will be 0.5mv2, but still the same number because nothing has been lost to friction.
So now looking at the horizontal surface, KE final will be zero, and KE initial will be equal to mgh...
That makes sense, except I'm going to have F*d on one side ... and to solve for F, I'll need acceleration (F=ma) ? I can't seem to see my way through that.
"According to the work-energy theorem, the net work on an object causes a change in the kinetic energy of the object. The formula for net work is net work = change in kinetic energy = final kinetic energy - initial kinetic energy."
But at some point I have to solve for displacement, so how can...
Homework Statement
A block slides down a frictionless incline (30 degrees above horizontal) for L=1.4m until it meets a horizontal surface with coefficient of kinetic friction 0.3 before coming to rest. Use work and energy to find the distance that the block slides on the horizontal surface...
Homework Statement
[/B]
The block, initially at rest, slides down the ramp and compresses the spring 0.03 m.
Theta = 30 degrees
L = 1.25 m
M of block = 2 kg
Δx = 0.03 m
1) Write the expression for the initial and final energy states
2) Find the spring constant K
Homework Equations
mgh...