I did an experiment on a force table with pulleys and weights. We were suppose to, through trial and error, find weights and add them on strings that were placed at different angles, so that the ring in the middle would become centered. Attached is the data table. I am suppose to calculate the...
A rocket is fired at an angle from the top of a tower of height 50.0m. Because of the designs of its engines, its position coordinates are of the form x(t)=A+Bt^2 and y(t)=C+Dt^3, where A, B, C, and D are constants. The acceleration of the rocket after 1.00s after firing is a=...
So I did an equilibrium experiment using pulleys and weights, where we set up different weights at different angles and had to get the right angle and weight to make the ring in the middle be centered. I am suppose to find the X and Y components of the forces. I am sort of confused on which...
From a position vs. time graph I calculated the slope of 2 tangent lines which is the velocity. The first slope was 390 cm/sec and the second slope was 486 cm/sec. Now to get the acceleration I thought I take difference of the velocities divided by the time interval 0.9 (1/60 sec) or 0.015sec...
A ball is thrown upward with speed V. At the instant the ball is thrown, second ball is dropped from rest at a height H above the thrown ball, but not directly above it. Ignore air resistance. How much time elapses before the balls pass each other? Find a value of H such that the ball thrown...
I have a position vs. time graph which is slightly curved. I found the slope of 2 tangent lines which I know are the velocity. My question is how do I get the acceleration using these 2 slopes. One slope is 295cm/s and the other is 575cm/s. I know that avg acceleration is final velocity -...
Ooo I think I get it. That means the slope should be twice the acceleration? So that means my acceleration from the slope would be 1000cm/s^2 which converts into 10m/s^2 which seems pretty close to 9.8m/s^2?
A sprinter accelerates to his max speed in 4.0s. He then maintains this speed for the remainder of a 100-m race, finishing with a total time of 9.1s. What is the runner's avg acceleration during the first 4.0s, during the last 5.1s?
I know that the runner starts at 0-m for his initial position...
Yes, its from an experiment I did in class. It was a free falling plummet ignoring air resistance. I basically got the slope from my line of best fit. So the y-value would be the velocity squared and the position would b the x-values. Dividing y/x would get the slope and since the y-value is in...
I asked this question a couple days ago but I am still confused. I have a velocity squared vs position graph and I am confused on what the slope means. I'm guessing that the slope is the displacement? I'm am also suppose to find the acceleration using this slope. The slope I calculated right now...
I calculated a slope from a best fit line of a velocity squared vs position graph but I don't understand what this slope represents. My units come out to be in cm/seconds squared, so does that mean that's my acceleration? It seems to be way off from 9.80m/sec^2 which should be my acceleration...
Consider the line L(t)=<4t-1,2+2t>. Then L intersects:
1. the x-axis at point ____ when t=____
2. the y-axis at point ____ when t=____
3. the parabola y=x^2 at the points _____ and _____ when t=_____ and t=______
I am confused on how to approach this problem. Do I just make x=4t-1 and y=2+2t?
I know that the area under an acceleration vs. time graph is the velocity, but how do I use that information to sketch a velocity vs. time graph. Is the area considered the slope for the velocity graph? Also how is it possible for a car to be slowing down during the first half of the motion...
Consider the two lines
L1: x=-2t y=1+2t z=3t and
L2: x=-9+5s y=36+2s z=1+5s
Find the point of intersection of the two lines.
My teacher said that I should use system of equations to solve for the point, but I am sort of confused on what to do because there are 2...