Recent content by N-Gin

The equations you mentioned are actually irrelevant and they don't help at all. What you really need is to use the basic definition of acceleration. First you find acceleration in x and y direction and then use Pythagorean theorem to find the magnitude (acceleration is a vector!).

Generally, polar coordinates can be converted to Cartesian using these equations: x=r\cos\theta y=r\sin\theta r^2=x^2+y^2 So, for the part b) you have r=2a\sin\theta r^2=2ay x^2+y^2-2ay=0 x^2+(y^2-2ay+a^2)=a^2 x^2+(y-a)^2=a^2 Now, the last equation is implicit...

x(t) is the function with t as a variable. It tells you what is the distance x from the traffic light at the time t.

If you know how far are the truck and the car from the traffic light at any moment, this problem is easy to solve. So, for the car you have x(t)=\frac{1}{2}at^2. And for the truck x(t)=x_0+vt. Now you just need to figure what is x_0 and what is the condition for the car and the...

First you need to know what is x in your equation. In this case it would be the distance from the bottom of the shaft to the bolt (at the certain time t). So, you actually want to know x_0, because it's the initial height. After time interval t, x is zero (bolt has reached the bottom). That's...

I'm not sure how exactly you want do find the relationship between the force and the speed of a tennis ball. I don't know if this will help, but I have a suggestion. The simplest method would probably be putting the ball on the instrument which can precisely measure the force exerted upon it and...

You sholud first find the distance between the two given objects as a function of time. I'm not sure if \theta in your picture is 90^{\circ} angle or not, but cosine rule will definitely work. Then you find the first derivative of that function in order to find the time at which the distance...

Yes, use Coulomb's law to calculate the forces. Just don't forget to treat them as vectors when calculating the net force. There's some basic trigonometry involved so it shouldn't be a problem.

You are almost done! You know the value of D, and by eliminating T from the first two equations you shold get the right answer.

It's best to assume that body is accelerating at a with initial speed v_{0} and height h_{0}. Maybe some assumptions aren't true, but we don't know that before we try to solve the problem. So, what you actually have is two equations of the form h=h_{0}-v_{0}t-\frac{1}{2}at^{2}. You are...

The best thing to do in such cases is to find the equations of motion for the each body, i.e. their exact location at any moment. Let the stones move along the y-axis (positive direction upwards). For stone thrown upwards, you have y_{up}=v_{0}t-\frac{1}{2}gt^{2}, where v_{0} is the...

You just need to use the Lorentz force. \vec{F}=q(\vec{E}+\vec{v}\times\vec{B}) Remember, \vec{E} and \vec{B} are the fields not produced by charge q.

What exactly don't you get? We can't help you if you don't show us your work! Generally, the first one involves Newton's laws (also look for the forces on the body). The second one involves some simple 2D kinematics (look for topics on projectile motion).

You must treat electric field as a vector! Draw the electric field vectors of each charge and with a little bit of trigonometry it should be no problem.

E = \frac {q}{4\pi\varepsilon(z-\frac{1}{2}d)^2} - \frac{q}{4\pi\varepsilon(z+\frac{1}{2}d)^2} E = \frac {q}{4\pi\varepsilon}[(z-\frac{1}{2}d)^{-2}} - (z+\frac{1}{2}d)^{-2}}] E = \frac {q}{4\pi\varepsilon}[(z(1-\frac{d}{2z}))^{-2}} - (z(1+\frac{d}{2z}))^{-2}}] E = \frac...