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Since Physics teachers are in very high demand, it is likely that you can get in through an...
You need to set up equations for each of the following:
1. The projectile motion problem that starts when the string is no longer taught.
2. The position (angle) at which the ball enters projectile motion (when the gravitational force inward surpasses the necessary centripetal force).
3. The...
Consider that a string could be used as a timing mechanism; as the propeller spins, the string gets pulled tighter (as it wraps around the axle.)
Another idea is a very well lubricated nut timing mechanism. Place a nut on the (threaded) axle, and don't allow it (the nut) to rotate. As the axle...
Try to think in terms of force acting on the ball.
The ball is moving downward, and then when it hits the water, what happens?
The ball is moving downward in the water, and then when it strikes the bottom, what happens?
Wikipedia is starting in the reference frame of the source of the waves in order to find the time between successive crests hitting the moving observer in the reference frame of the source.
Then Wikipedia uses this time in the time dilation equation in the observer's reference frame in order...
Do not solve for the left side of the catapult arm separately from the right. Once you have the moment of inertia for the whole arm, use that to find the angular velocity of the whole arm.
As for the energy equation you have written, at the beginning, the catapult only has spring potential...
I was thinking that Unillusive could use the equations for the interaction between charged particles, and discuss any similarities to the equation for the interaction between magnets.
I.e. the inverse square relationship, and as you said, the fact that like charges attract, while unlike...
Good, this is correct. You may, however, want to reference the following formula in your explanation:
\stackrel{\rightarrow}{F}=\stackrel{\rightarrow}{I}L\times\stackrel{\rightarrow}{B}
Here you are supposed to be comparing the electric force between two charged particles to the magnetic...
I would recommend against force dynamics. I think the problem would be much easier for you to solve using energy dynamics.
The formula you would use to do this is:
KE_{arm}=\frac{1}{2}I\omega^2
Now, this next thing is important:
Are you throwing the projectile with a sling, or is it...
Draw the cone on your paper as a triangle facing down. Pick some point along the center axis. The radius is the distance from the center to the edge, and the height is the distance from the bottom to the current location.
Now you can use trigonometry to solve for the radius in terms of the...
If it is fixed at one end and free at the other, one end must be a node, and the other end must be an antinode.
Try to draw what the acceptable standing waves would look like; that will help you to find their wavelengths.
Don't think of the start of this problem, think of the problem when the tank is at equilibrium. If the tank is to be in equilibrium, the velocity of water entering times the cross sectional area of the entranceway is equal to the velocity of water leaving times the cross sectional area of the...
We is the rate of water entering the tank; it is equal to the velocity of water entering the tank times the cross-sectional area.
Wl is the rate of water leaving the tank; it is equal to the velocity of water leaving the tank times the cross-sectional area. So you start this with bernouli's...
Re: Please help me with this Question based on angular momentum.....
The moment of inertia of this body should be the moment of inertia of the uniform rod about the axis plus the moment of inertia of the bullet.
Think of the bullet as a point particle of mass m embedded in the center of...
Velocity is defined as the rate of change of position divided by the rate of change of time.
This is the same as saying that velocity is the slope of the position vs. time graph.
Therefore, instantaneous velocity at 1s is the slope of the position vs. time graph at t=1s.
In boat A's reference frame, boat B's velocity is Vbj, so what is boat B's velocity in the lake's reference frame?
Now do the same thing for the missile.
Once you have the missile's velocity in the lake's reference frame, you can solve the problem like a projectile motion problem.
Start by writing an equation for the rate of change of water in the tank.
i.e. \frac{dW}{dt} = W_e - W_l
where W is the amount of water in the tank, We is the amount of water entering the tank, and Wl is the amount of water leaving the tank.
When the 2kg mass reaches its maximum speed, its acceleration should be zero.
If the 2kg mass's acceleration is zero, the net force acting on it should be zero.
The easiest approach to this problem, is therefore to ignore the initial conditions, and simply solve for the location, x, where...
Try using:
cos(\theta)=\frac{x}{r}
where x is the distance of the 2kg mass from the upright, and r is the amount of string between the 2kg mass and the pulley.
Then, you should be able to determine r as a function of l, and x as a function of r and y.
I am going to assume that you are looking for the instantaneous acceleration of the 2kg mass. In order to find this, use:
\sum F = ma = 2a
and use the fact that the tension in the string can be calculated at the 1kg mass to be:
T=mg-ma=g-a
Now consider all of the forces acting on the...
Looks correct to me.
It may help you to think of these situations in terms of "phasors"
http://en.wikipedia.org/wiki/Phasor
Draw a vector representing each wave with a length equal to the amplitude, and an angle equal to the phase. The resultant vector will give you the amplitude and phase of...
You solved the problem correctly!
It was good of you to second guess your answer when you found a number larger than any you have seen before, however, when the time for acceleration is that small, and the change in velocity is not, the acceleration MUST be very large.
As a general rule of...
Try drawing a right triangle from here to the center of the moon to the edge of the moon, and back.
One leg of the triangle will be the distance from here to the moon, and the other leg will be the radius of the moon. The angle between the long leg and the hypotenuse will be half of the...