I'm not sure, but if you say that the tension of the dashpot is Z_d*(x_1 - x_2) , then this is the magnitude of the force the dashpot exerts on the strings. Using Newtons third law, the strings will exert an opposite and equal force on the dashpot. Thus, we would have Z_d*(x_1 - x_2) =...
I am currently trying out for the US Physics team, and am working on some practice exams. I was wondering if anyone could offer some assistance on the following problems below:
Problem A2
http://www.compadre.org/psrc/evals/IPO_Exam_2_2003.pdf
Problem's A4 and B1...
The wording in your description may be slightly confusing to posters. Perhaps if you could include a diagram of the problem. Anyways...when you work with statics...always remember that about any point, the torques always = 0. Also, sum of the x forces = 0 and sum of the y forces = 0.
Net work done on object = (horizontal component of 118 - frictional force) * distance
Net work done on object = change in kinetic energy (work-energy theorem)
The rope must also overcome the force of gravity acting on Tarzan when he is at the bottom of the arc.
Thusly, the tension of the rope = Fcent + Fgrav at this point.
The question is asking for "when", which implies time. THus, take the derivatives, and solve for t when the derivative = 0. AFter you have found your values for t, then you will need to test whether the value is a maximum or a minimum.
(if you don't find a max or min. with these values...
...x = v cos (theta) t
divide both sides by vcos(theta)...and you get
t = x/(v cos (theta)
Now plug this into the equation for y...
When you do this, you should be able to solve for initial velocity.
To start off...
Gravitational Potential Energy = Linear Kinetic Energy + Rotational Kinetic Energy
...Rotational Kinetic Energy = 1/2 I w^2
w(its called omega) = v/r
I of Sphere = (2/5)mr^2...
..thusly...Rotational Kinetic Energy = (1/2)[(2/5)mr^2][v/r]^2...which simplifies to...
Tarzan is moving in an arc...that means that centripetal acceleration is involved...(since he's moving in a circular direction).
Thusly, there is a centripetal force acting on Tarzan (Fc = mv^2/R) and the graviational force as well (F = mg)...basically...work with that info...and also...try...
P = F/A;
Change in momentum = Impulse = Force * time
Change in energy = Work = (Force)(distance)(cos (theta))...since this a rocket..the force will most likely be in the same direction as the movement...so...cos(0) = 1...and you have...
Work = Force(distance).
Or, since you know the...
...think...when the car is at rest...what forces are acting on it? What forces are acting on the ball? Is the ball in equilibrium?
...when the car is accelerated up the incline...how does this acceleration affect the ball (remember, princaple of equivalence?)
yeah...sorry 'bout that..I left out the sumation part... but basically follow Spectre5's advice...free body diagrams are essential to solving most force related Physics problems
F = ma
Thus, when the car is accelerating, the force acting on the car not only must overcome the force of gravity, but it also must accelerate the car.
Looks like you're using mr^2 for the moment of Inertia...which would work if it were a ring..
..however, the moment of Inertia of a thick solid disk is:
I = 0.5mr^2..
...hope that helps!
The time of explosion is when the vertical velocity = 0 (since it is at the top of the path). Use this info when calculating...
part B just involves finding the kinetic energy of the object before it explodes, which is:
KE= 1/2mv^2
THe KE afterwards would just be the same thing, except...
Now, remember...this is minumum speed...
...so the tension of the string is mimumum, or zero when the ball is at the top of its path (thus, only the force of gravity would equal the centripetal force.)
With this knowledge, draw a free body diagram of the ball when it is on the top of the...
Well...the method I just gave you is based on classical mechanics, which basically works ok for anything << speed of light. However, anything remotely close to the speed of light (such as something with a speed of 0.1c)...would require the relativistic definitions...
In classical mechanics...
k...in this case...the force of gravity = centripital force...which I think you have figured out already...
You are correct in stating that Centripital force= mv^2 / R
Now draw a free body diagram to find the force of gravity...its not mg tan (angle)...its mg * something...
...work from...
Yes. You can find the velocity of an electron accelerated through a particular potential difference.
Remember that:
potential differerence (aka voltage) * charge = Change in potential energy (U)...
Since you know the charge of an electron...you can simply multiply this by the...
Remember that A(centripital) = v^2 / R...
...so your answer to problem 1 looks correct.
On problem 2, how did you go from a velocity of 48.1 m/s to that of 4.8 m/s? The mass would just be the pilots mass...because its asking for the centripital force acting on the pilot.
Try to draw a free body diagram. In one direction, you have 722.1 Newtons acting on the piano. In the opposing direction, you have the force of gravity.
F = ma
Mover's force - force of gravity (sin 9.1 * mg) = Net Force
Acceleration = Net Force/Mass
...with this acceleration, just...
Work = Force * distance
Remember, you only need to multiply stuff by the cos (angle) or sin (angle) to get that component of force so it matches up with movement.
In this case, the 60 N applied in the opposite direction as the movement...so you don't need to find any of its components...
First, you mentioned the distance between B and C as 150 m, and later said it was 1500.
A second thing to keep in watch, when you converted m to km, and multiplied it by cos(45) or sin(45), you changed distance to velocity by adding the "/h". Really, 1.5km * cos(45) = 1.06 km
Keep in mind...
Uhhhmm...there is no need for the quadratic equation on this problem. It can be solved by using the general kinematic equation in most Physics textbooks.
Since the stone is at rest and has no initial velocity...we can use:
distance = (1/2)at^2
We know the distance the stone travels...
I think Moore only cares how well the film does before the upcomming November election. Since the oscars will be afterwards, I think whether his film is nominated or not will probably not bother him.
I think the 10 dimensional universe theory (4 macroscopic dimensions, and 6 microscopic dimensions) has to do with string theory, which has not yet been proven experimentally (Which would be rather difficult).
I suppose the latter explanation is how general relativity would explain it. I...
There are 6.022 * 10^23 atoms in 1 mole.
Molar mass of hydrogen is around 1 g/mole...carbon is around 12 g/mole...and oxygen is about 16 g/mole...
You have 12,000 grams...you can find the moles and then thus find the atoms. :-D
Blackmamba, write an equation describing the distance between the bear and tourist, and make it relative to time:
distance between bear and car = 20 m -(bear speed * time - runner speed * time)...work from there to find the time.
With time, and runner speed, you can find distance...
No...s^2 is simply a unit. It is NOT a variable. You are taking the derivative of the function with respect to the variable t, so try to envision that particular term as 2.20t^2...and the derivative of that would be 4.40t.
yeah...i know. ;-D Most people probably would solve this problem like you did...but I thought it would be interesting to post a method that doesn't involve a big 5 equation derived from calculus...and uses, in a sense, basic mathematics to find acceleration.
Needpersonhelp's method works, largely cause in this case the acceleration is constant. Another method that will work is outlined below:
14.4 m
25.6 m 2 s later
40 m Another 2 s later
57.6 m Another 2 s later
Now, find the differences in the distances...and you...
Integrating a function is a method of finding the area beneath it. There's an elegant proof found in most calculus books that connects Rieman's sums (which is really just adding a bunch of infinately small shapes together to find area) to integration. Thus, since integration is much simpler...
actually, based on your answer of a = 4.8 ft/sec^2...it appears that you must have used the above method, or a similar one...but instead of having the length be equal to 300 ft (100 yards)...looks like you had it equal to 100 ft. So anyways, I'm willing to guess that you had your method...
david90, here's a way you can solve it algebraicly:
For the first 2.5 seconds, the sprinter is accelerating...thus...we can describe that part as
distance = (1/2)at^2, with t = 2.5 it would be = (1/2)a(2.5)^2
For the last 7.1 seconds, we know that the speed is both constant and...
The stones are thrown at the same horizontal speed, but the vertical heights of the buildings are different. The time it takes for the stone to hit the floor has nothing to do with horizontal speed (now we are neglecting air resistance here), but only to do with the height with which it was...
Galileo is correct with his answer (isn't he always correct...i mean...about moons and Jupiter...acceleration on Earth independent of mass...except maybe for his non-relativistic work on motion for frame of references). Anyways, the answer to the problem is that the angle = arctan u. If the...
yeah...knowledge of power...you calculations look fine to me...as force is a vector, so a negative value should simply mean that it is opposite to the motion of the car.
You could also solve this way:
let t = time it takes to fall.
since x = .5gt^2 (one of the general motion equations)
h = .5g(t^2)
.75h = .5g(t - 1)^2 (since time will be t-1 when it has fallen .75h)
just subtract one equation from another to get .25h = .5(t^2) - .5(t - 1)^2...and...
Princexcharles...1000 ms is equal to 1 second...so 30 ms would be 0.03 seconds.
Also, Physicsneededhelp...for future reference, make sure you check your units before you go through with your calculations. In your earlier post, you were dividing meters/seconds by miliseconds...make sure all...
Yeah...everyone above has given some pretty good advice...I just want to emphasize that it is VERY important to do practice problems...the more problems to do...the better situation you'll be in.
terpsgirl...for the first problem...use the equation that states that impulse = change in momentum...the equation is:
Force * time = change in velocity * mass.
Think of what the original velocity is, and the final velocity...and u should be able to find the change in the velocity...