Recent content by candycooke

I need to know the non-uniform acceleration equation to determine final position given: initial position initial velocity initial acceleration jerk time I know that the equation with uniform acceleration is: x(f) = x(i) + v(i)t + (1/2)at^2 I think the equation I'm looking for is the...

The position y of a particle moving along the y-axis depends on the time t according to the equation y = at - bt2. The dimensions of the quantities a and b are respectively: A. L2/T, L3/T2 B. L/T2, L2/T C. L/T, L/T2 D. L3/T, T2/L E. none of these I don't understand what the question is...

A 12.5 g bullet is shot into a ballistic pendulum that has a mass of 2.37 kg. The pendulum rises a distance of 9.55 cm above its resting position. What was the speed of that bullet? m(bullet) = 0.0125 kg m(pendulum) = 2.37 kg h = 0.0955 m I know that I'm supposed to use conservation of...

Thank you for your help Doc Al. It all makes sense now.

m2 = [m1gsin(theta) + (mu)m1gcos(theta)] / g m2 = [m1g[sin(theta) + (mu)cos(theta)]]/g m2 = m1[sin(theta) + (mu)cos(theta)] So factoring has eliminated g but m1 and m2 are sill part of the equation, leaving me just as confused as before.

0 = m2g - m1gsin(theta) - (mu)m1gcos(theta) m2g = m1gsin(theta) + (mu)m1gcos(theta) m2 = [m1gsin(theta) + (mu)m1gcos(theta)] / g Although the equation makes a bit more sense, I'm still confused about how to give a numerical value for (mu) and (theta) so that m2/m1 > or < 1 without also...

a) Find a general expression for mass m2 so that m1 will move at a constant speed up the ramp. Your answer will be based on (mu) and (theta). b) Give a numerical value for (mu) and (theta) so that m2/m1 > 1 c) Give a numerical value for (mu) and (theta) so that m2/m1 < 1 Fg2 = m2g Fg1x...

A 3 kg mass is moving along a frictioless surface at 17 m/s. It then reaches a slope and starts up it. The slope has an angle of 10º and a small amount of friction. What is the coefficient of friction on the slope if the block slides 4.5 m up the slope before stopping? m=3 kg v(i)=17 m/s...

Since v_fy=0 at the highest point: t/2=(0-25sin\theta)/-9.8m/s^2 therefore: t=2(0-25sin\theta)/-9.8m/s^2 but this is not equal to t=30m/(25cos\theta) I am so confused.

since the porabola is symetrical, than v(i)=v(f) for y components: t=v(f)-v(i)a =0a for x components: t=d/v =30m/25cos\theta therefore:0=30m/25cos\theta \theta=33.6 So I've figured out one angle but I am unsure how to determine the second possible angle.

An arrow is shot with an initial speed of 25 m/s (and lands at the same height as its launch height). There are two possible angles at which it chould be aimed so that it goes 30 meters. What are they? Givens: v(i)=25m/s a=-9.8m/s^2 d(x)=30m I know the first step is to determine the...

A person who is properly constrained by an over-the-shoulder seat belt has a good chance of surviving a car collision if the deceleration does not exceed 30 "g's" (1.00 g = 9.80 m/s^2). Assuming uniform deceleration at this rate, calculate the distance over which the front end of the car must...