Recent content by Rawrr!

How about using the conservation of momentum to work it out? Split the momentum into individual components, in the x and y directions, and work from there. It also may help you visualize the situation if you draw the scenario out, both initial conditions, and final conditions.

To help start you off to better understand, let's go around the loop from Y→D→A→X→Y We'll hit the resistor of 820Ω. Since we already defined the current I1 traveling through this part of the wire, V820Ω = I1(820Ω) now following this along clockwise V820Ω + V5V + V180Ω + V680Ω +...

It will probably be easiest for you to first designate your currents through the loops. Which, you can choose arbitrarily. So, to start you off, I would compel you to glance at the end of page 3 in your attached pdf, to give you something to look at, when taking into account Kirchoff's second...

As gneill put it, it helps to split them up, and get individual equations. I would split the circuit up into three different loops. loop AXYD loop XBCY loop ABCD And write out the equations in terms of voltages. V=IR. Also, don't forget to take into account that your current will split or...

Mechanical Energy is the sum of the potential energy, and the kinetic energy. Emechanical = Ekinetic + EPotential Since you're not close to the surface of the Earth, equating the potential energy to (mgh) is not applicable. Epotential = (-GMm)/r - r is the distance from the...

A is very exothermic. Oxygen and hydrogen are often used together as a means of propulsion in rockets. (Although to save space, they're usually condensed into a liquid state). It's a synthesis reaction as you're turning two reactants into one product.

Where you have i, j , and k, that remain constant along the axes, polar coordinates are just another way of addressing the same problem of denoting a position in space. Suppose, in just using the xy-plane, you have a circle of radius one, centered at the origin. Using Cartesian coordinates...

In the initial equation of the vector r, you can see that there are given unit vectors, i, and unit vector j . These are simply denoting a direction in Cartesian coordinates. Because a unit vector is equal to one, it's only there to designate movement along a direction, as multiplying by one...

Looks correct! Just remember that the velocity of the gun is in the opposite direction of the ball. Just like when you shoot a bullet, the bullet goes one way, and the gun goes in the other (the recoil). (If it's going to be graded, it might be beneficial to write down the direction of the...

The final velocity would be the velocity it has when it leaves the barrel. So in your initial question, you say the ball leaves the gun with a velocity of 45 m/s . The initial velocity of the ball would be zero as both the gun and the ball are at rest, before the gun is shot.

Here, you would use Newton's third law, and that the forces should equal. "For every action, there is an equal, but opposite reaction." The paintball gun forces the paintball in the +x direction, the ball pushes the gun in the -x direction. (this is why you feel recoil when shooting). Find the...

Yes, that is correct. Since v0 is split by the angle of 45 degrees in its initial conditions, the x component, as you stated, is v0*(cos(45)), and the y component would be V0*(sin(45)). While the y velocity changes due to gravity, the x velocity remains the same. At max height, the y...

To find the resulting Force, break down each force into its separate components. the force along x, the force along y, the force along z. now simply add these together. Fx = Fx,1 + Fx,2 Fy = Fy,1 + Fy,2 Fz = Fz,1 + Fz,2 From these, you should get the answer. (38.22x + 67.99y +...

In 32a) to find the distance, you seemed to have used the mass of the car, in place of its velocity. That being said, i think your equation could be modified. The question is relating to the distance between the car and truck while in motion. the velocity remains constant. distance =...

In your attachment, the answer is (38.22i + 67.99j + 17.5k). In three dimensional vectors, i, j, and k are simply used to describe the axes. i=x , j=y, k=z . So you have a magnitude of 38.22 along the x-axis, 67.99 along the y-axis, and 17.5 along the z-axis. It's very similar to 2-d vectors...