You've got the size of the force down. As for the direction: you have opposite charges, which means they attract one another. So your first answer (the negative one) is the one you should use.
For the first part, you know that the velocity is given by the wavelength multiplied by the frequency. Once you have the velocity, you multiply it with the time interval to get the distance.
I'm confused about what the question is referring to in the second part.
Well let's start off simple. We know that 2*pi*r is the circumference of a circle. So the circumference of a circle multiplied by a height h means that we now the have the surface area of a cylinder (excluding tops). Do you see that at least? Once we have the surface area, it's just one more...
You would need to use the centripetal force equation. If the radius was four times smaller, the force needed to keep the bull in orbit would be << answer deleted by berkeman >> times bigger.
The electric field inside a conductor must be zero, so there can be no charge inside the sphere. The remaining charges array themselves on the surface until there is no field inside the sphere, and the only way that can work is if they are concentrated uniformly on that surface.
Yes...but the net work is only equal to the change in kinetic energy. What your book is saying (spring's potential energy) has nothing to do with what you're saying. For a conservative force (ie. only a position-dependent force) like Hooke's law, the force will equal the negative gradient of the...
The first part is wrong. You need to use the following formula:
Work = Force * Distance
So it should just be 354/47, giving you 7.5 meters.
I can't tell from what you've written if you're right on part 2 or not, so here's how you do it. You need to set the kinetic energy of the bullet...
The answer is no. The force exerted by your hand was necessary to cause a change in motion, but not necessary for motion itself. This point was the fundamental difference between Newtonian and Aristotelian mechanics.
Please at least attempt a solution next time. I'll help you out anyway because I'm bored.
The basic strategy behind solving this problem centers on conservation of momentum. Before we get to that, however, we need to find out the mass of the student. We know the student's weight, W, and from...
The first objective is to determine the car's deceleration. After that, we use kinematics to calculate the car's initial velocity and compare it to the speed limit.
The frictional force acting against the car's motion can be written as:
F_f = u_kF_N
The normal force is mg, so the...
Ok this is a basic Atwood machine problem. There are many ways to solve it. I'll do it using the Lagrangian formulation since it's asking for energy methods, but you can do it through standard Newtonian mechanics as well (the two formalisms are physically equivalent). I'll assume that the mass...
It would be appropriate to use it if you're given an equation for some EM wave. Then the constant in front squared would represent the intensity of the wave.
This discussion is getting too cluttered when you're already on the right path. The only mistake you made here is that you got the height wrong. It should be 26.7 - 18 = 8.7 meters. Remember, you want the height difference between your original point and your final point (so if radius is 9, the...
This question seems confusing. Are you saying something to the effect of...the Earth can't spin too fast because its intrinsic angular momentum would have to equal its orbital angular momentum?
Wait this is weird...is this a double post or something? I already answered this question at this other thread created by the same user: https://www.physicsforums.com/showthread.php?t=359872"
For Part i, use the equation y = (1/2)gt^2, where y is the height and t is the time. Solve for t. You should get 1.43 seconds. To find out how fast it's moving, use the equation v = sqrt(2gh), where v is the speed and h is the height. You should get 14 meters per second.
For part ii, if it...
The total velocity can be written as components of the velocity in the x direction and in the y direction:
V(total) = sqrt(V(x)^2 + V(y)^2)
where V(x) = V(0)cos(theta) and V(y) = V(0)sin(theta)
In the x direction, the object travels the following distance (let's call it D(x), which is...
You will need the range equation:
R = v^2 * sin(2*theta) / g. Knowing theta and R (the range), solve for v.
The time in the air is given by t = 2*v*sin(theta)/g. You know v and theta, so solve for t. If you need me to derive these equations, let me know.
This seems to be an Atwood machine problem, although I don't understand the last part of the problem statement.
You will want to set up the equations of motion for each mass. For mass m(1):
F1 = m(1)*a = T -m(1)*g = 0
For m(2):
F2 = m(2)*a = m(2)*g - T = 0
To solve for tension...