# Lab Report

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Hello , I will send u guys the material about it. The that we have done + the work sheet etc.

I want it at least two pages , it is worth 100 point

Physics I: Force Worksheet Class: Section: Lab Group: Assignment Date: UP1 001 32 9-26-17 Names: James Fountain IV Abhishek Kumbhani Shabib Alzuabi answer key: numbers Table 1: Determining Angle of Track 14.5 Rise, m Hypoteneuse, m Angle from Horizontal , degrees Table 2: Measurements of Cart Mass of Cart, kg 186 Experimental Force, N 4.47 Theoretical Force, N % Error of Forces, % Theoretical Acceleration, m/s^2 Acceleration from Position, m/s^2 Acceleration from Velocity, m/s^2 %Error between Theoretical and Nearest Measured Acceleration et James Fountain IV Abhishek Kumbhani Shabib Alzuabi text data rements of Cart 0.517 .380 .394 3,553299 3.55 .764 .926 .921 17.0 17,04669 Free Body Diagrams Instructions : Present free body diagrams of the cart, one when tied back and one when freely r indicate the coordinate system; label and show the direction of all forces and off-axis components; a diagram. From the resulting set of four equations, derive formula in a readable, understandable, lo cart, and 2) the expected acceleration on the rolling 1. Diagram expected tension James 2. Diagram of expected James Shabib Shabib Abhishek Abhishek Diagrams nd one when freely rolling, from each student. For each diagram, clearly -axis components; and write a sum of forces equation for each axis from the e, understandable, logical way for 1) the expected tension on the restrained eration on the rolling cart. 2. Diagram of expected acceleration Graphs of Cart in Motion Figure 1: Position of Cart over Time Position vs. Time 0,400 0,350 Position (m) 0,300 0,250 0,200 0,150 0,100 0,050 0,000 0,150 0,170 0,190 0,210 0,230 0,250 0,270 Time (s) Figure 2: Velocity of Cart over Time with Linea Velocity vs. Time 0,800 0,700 0,700 Velocity (m/s) 0,600 0,500 0,400 0,300 0,200 0,100 0,000 0,150 0,170 0,190 0,210 0,230 0,250 Time (s) 0,270 0,290 art in Motion of Cart over Time y = 0,4631x2 + 0,4116x + 0,1513 R² = 1 0,270 0,290 0,310 0,330 over Time with Linear Fit 0,350 y = 0,9214x + 0,4133 R² = 0,9998 0,290 0,310 0,330 0,350 Motion Data Instructions : Put all raw data in lefthand columns. In the columns on the right, place (at the top) just the portion of your data that you are using in your analysis. Experimental Data Selected Data time (s) position (m) velocity (m/s) time (s) position (m) 0.033333 0.066666 0.099999 0.133332 0.166665 0.199998 0.233331 0.266664 0.299997 0.33333 0.366663 0.399996 0.433329 0.466662 0.499995 0.533328 0.566661 0.599994 0.633327 0.66666 0.699993 0.733326 0.766659 0.799992 0.833325 0.866658 0.899991 0.933324 0.966657 0.99999 0.1761305 0.1799035 0.1963675 0.214032 0.2327255 0.252105 0.2725135 0.293951 0.3164175 0.339913 0.3644375 0.3898195 0.418117 0.442813 0.470253 0.499408 0.5297635 0.560805 0.593047 0.628033 0.6621615 0.6971475 0.733334 0.7698635 0.8075935 0.8463525 0.8861405 0.9261 0.96726 1.009792 0.22095137618 0.348863071964 0.473430484305 0.533513251799 0.570957792911 0.598255149218 0.627982113154 0.658566585666 0.689293976273 0.721164711647 0.752606692734 0.788050380504 0.794910449104 0.802770944376 0.846075127418 0.887092620926 0.921821718217 0.956122061221 0.999997916646 1.02886737201 1.04330209969 1.06716942169 1.0908938256 1.11661908286 1.14648896489 1.17464382977 1.19722488892 1.21920569206 1.2441734834 1.26253845872 0.199998 0.233331 0.266664 0.299997 0.33333 0.252105 0.2725135 0.293951 0.3164175 0.339913 t, place (at the top) just the s. ed Data velocity (m/s) 0.598255149218 0.627982113154 0.658566585666 0.689293976273 0.721164711647 0,200 0,233 0,267 0,300 0,333 0.252105 0.2725135 0.293951 0.3164175 0.339913 0,200 0,233 0,267 0,300 0,333 0,200 0,233 0,267 0,300 0,333 0,252 0,273 0,294 0,316 0,340 0,598 0,628 0,659 0,689 0,721 Position vs. Time 0,400 0,350 Position (m) 0,300 0,250 y = 0,4631x2 + 0,4116x + 0,1513 R² = 1 0,200 0,150 0,100 0,050 0,000 0,150 0,170 0,190 0,210 0,230 0,250 0,270 0,290 0,310 0,330 Time (s) Velocity vs. Time 0,800 Velocity (m/s) 0,700 0,600 y = 0,9214x + 0,4133 R² = 0,9998 0,500 0,400 0,300 0,200 0,100 0,000 0,150 0,170 0,190 0,210 0,230 0,250 Time (s) 0,270 0,290 0,310 0,330 0,4116x + 0,1513 0,330 0,350 ,9214x + 0,4133 0,330 0,350
Physics I: Force Worksheet Class: Section: Lab Group: Assignment Date: Names: answer key: numbers Table 1: Determining Angle of Track Rise, m Hypoteneuse, m Angle from Horizontal , degrees Table 2: Measurements of Cart Mass of Cart, kg Experimental Force, N Theoretical Force, N % Error of Forces, % Theoretical Acceleration, m/s^2 Acceleration from Position, m/s^2 Acceleration from Velocity, m/s^2 %Error between Theoretical and Nearest Measured Acceleration et text rements of Cart data Free Body Diagrams Instructions : Present free body diagrams of the cart, one when tied back and one when freely r indicate the coordinate system; label and show the direction of all forces and off-axis components; a diagram. From the resulting set of four equations, derive formula in a readable, understandable, lo cart, and 2) the expected acceleration on the rolling Diagrams nd one when freely rolling, from each student. For each diagram, clearly -axis components; and write a sum of forces equation for each axis from the e, understandable, logical way for 1) the expected tension on the restrained eration on the rolling cart. Graphs of Cart in Motion Figure 1: Position of Cart over Time Figure 2: Velocity of Cart over Time with Linea art in Motion of Cart over Time over Time with Linear Fit Motion Data Instructions : Put all raw data in lefthand columns. In the columns on the right, place (at the top) just the portion of your data that you are using in your analysis. Experimental Data time (s) position (m) velocity (m/s) Selected Data time (s) position (m) t, place (at the top) just the s. ed Data velocity (m/s)

JesseCraig
School: University of Maryland

I have done your work. Thank you.

1

Physics Lab Report
Student’s name:
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Physics Lab Report

2

Introduction/theory
An object can change its motion, direction or position under an application of a force.
The applied force is normally given by:
F = ma
Where F is the applied force, m is the mass of an object & a is the acceleration that an
object undergoes.
The summation of forces experienced by an object must be equal to zero for an object to
remain at equilibrium i.e. ∑ 𝐹x = 0 & ∑ 𝐹y = 0. This implies that the summation of both
horizontal and vertical forces must be equal ...

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Anonymous
Excellent job

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