Newtons Second Law-Acceleration
Mark Tomaszewski
Partner: Samantha Hawes
TA: Felix Kling
Physics 181 Section 17
Submitted on September 27th 2012
Abstract
The goal of this experiment was to apply Newtons 2nd law of motion (Force=ma)
to an apparatus, which
Range Vs. Height
Mark Tomaszewski
Partner: Samantha Hawes
TA: Felix Kling
Physics 181 Section 19
Submitted on September 13th 2012
Received Online: Wednesday, September 12, 2012 10:24 PM MST
Org Unit: PHYS181 FA12 001-019
File(s): Range Vs Height Lab Repor
Newtons Second Law
Lab report
Zach Capozzi
Lab partners:
Course PHY181-001
TA: Jeremy Long
Due date 8:00am on 10/1/2015
Abstract
The goal of the experiment was to find out the acceleration of the car on the track
as a function of both its applied force an
Range Vs. Height
Lab Report
Zach Capozzi
Lab Partner:
Course PHYS181-001
TA: Jeremy Long
Due date 8:00AM on 9/17/2015
Abstract
We measured the average range of a small silver ball being dropped from an
initial height with three different methods, then fro
Conservation of Momentum
Mark Tomaszewski
Partner: Samantha Hawes
TA: Felix Kling
Physics 181 Section 17
Submitted on October 18 2012
Submitted Online: October 17 2012
Abstract
For this experiment we set up a bunch of different trials where we had 2 carts
Friction
Mark Tomaszewski
Partner: Samantha Hawes
TA: Felix Kling
Physics 181 Section 17
Submitted on October 11 2012
Submitted Online: October 11 2012
Abstract
For this experiment there were 3 parts: The first part was to find the s for both the cork
sur
Energy Conservation
Lab Report
Zach Capozzi
Lab Partners: Jenelle & Caroline
Course PHY181-001
TA: Jeremy Long
Due Date: 8:00am on 10/15/2015
Abstract:
The goal of the experiment was to measure the velocity of the car on the roller
coaster track as it pas
Classical Mechanics
Lecture 12: Work and Energy
Todays Concepts:
Work & Kine6c Energy
Mechanics Lecture 7, Slide 1
Stuff you asked about:
I have no concept whatsoever of whats happening
This stu is geGng preHy hard. Having a lot of formulas without
Jason Nuss Physics 181 Baker 10/12/2007
Friction
Goals of the Experiment: The experiment will test the effect of the change in normal force on the change of frictional force by using known masses suspended vertically and known masses moving horizont
Standing Waves
Zachary Capozzi
Lab Partners: Jenelle & Carolina
Course: PHY181-001
TA: Jeremy Long
Due Date: 8:00am on 11/12/2015
Abstract
The lab performed this week was to have us better understand the principles of
physics already known through previou
Physics 181
Worksheets
Course Coordinator: Dr Haar
This Laboratory Manual
was produced for the
PHYSICS DEPARTMENT
UNIVERSITY OF ARIZONA
By
Brian LeRoy
2013
Based in part on previous
Physics Department
Laboratory Manuals
Copyright 2013
All rights reserved
1.41:
The total northward displacement is 3.25 km 1.50 km 1.75 km, , and the total westward displacement is 4.75 km . The magnitude of the net displacement is
1.75 km
2
4.75 km
2
5.06 km. The south and west displacements are the same, so
The d
1.35: A; Ax = (12.0 m ) sin 37.0 = 7.2 m, Ay = (12.0 m ) cos 37.0 = 9.6 m.
B; B x = (15.0 m ) cos 40.0 = 11.5 m, B y = −(15.0 m ) sin 40.0 = −9.6 m.
C ; C x = −( 6.0 m ) cos 60.0 = −3.0 m, C y = −( 6.0 m ) sin 60.0 = −5.2 m.
1.42: a) The x- and y-components of the sum are 1.30 cm + 4.10 cm = 5.40 cm, 2.25 cm + (3.75 cm) = 1.50 cm. b) Using Equations (1-8) and (1-9),
(5.4 0 cm) 2 ( 1.50 cm) 2 = 5.60 cm, arctan
1.50 5.40
= 344.5o ccw.
c) Similarly, 4.10 cm (1.30 cm) =
1.39: Using components as a check for any graphical method, the components of B are 12 m . Bx 14 .4 m and By 10 .8 m, A has one component, Ax
a) The x - and y - components of the sum are 2.4 m and 10.8 m, for a magnitude 10.8 2 2 77 .6 . of 2.4 m 1
1.25: The shape of the pile is not given, but gold coins stacked in a pile might well be in the shape of a pyramid, say with a height of 2 m and a base 3 m 3 m . The volume of such a pile is 6 m3 , and the calculations of Example 1-4 indicate that th
1.26: The surface area of the earth is about 4 R 2 5 10 14 m 2 , where R is the radius of the earth, about 6 10 6 m , so the surface area of all the oceans is about 4 10 14 m 2 . An average depth of about 10 km gives a volume of 4 1018 m3 4 10 24 cm3
1.45:
A B C
^ 12 .0 m sin 37.0 i ^ 15.0 m cos 40.0 i
12 .0 m cos 37.0 ^ j ^ 15.0 m sin 40.0 j
^ 7.2 m i
9.6 m ^ j
^ 6.0 m cos 60.0 i
6.0 m sin 60.0 ^ j
^ 11.5 m i ^ 3.0 m i
9.6 m ^ j 5.2 m ^ j
1.40: Using Equations (1.8) and (1.9), the magnitude and direction of each of the given vectors is: a) o 180 31.2o). b) c) o 360 19.2o).
( 8.6 cm) 2
(5.20 cm) 2 = 10.0 cm, arctan
5.20 8.60
= 148.8o (which is
( 9.7 m) 2 (7.75 km) 2
( 2.45 m) 2
1.37: Take the +x-direction to be forward and the +y-direction to be upward. Then the second force has components F2 x F2 cos 32 .4 433 N and F2 y F2 sin 32.4 275 N. The first force has components F1x
Fx F1x F2 x 1158 N and Fy
F1 y 725 N and F1 y F2
1.36:
(a)
tan
Ay AX
1.00 m 2.00 m
0.500
(b)
tan
0.500 360 26.6 333 Ay 1.00 m tan 0.500 Ax 2.00 m
1
(c ) (d) tan tan
1
tan Ay Ax
1
0.500 1.00 m 2.00 m 180
26.6 0.500 26.6 0.500 207 153
0.500 Ay Ax 0.500
tan tan
1
1.00 m 2.0
1.46: a) b)
A B C
^ 3.60 m cos 70 .0 i ^ 2.40 m cos 30.0 i 3.00 A
3.60 m sin 70 .0 ^ j 2.40 m sin 30.0 ^ j
^ 1.23 m i
3.38 m ^ j ^ 2.08 m i 1.20 m ^ j
4.00 B ^ 3.00 1.23 m i 3.00 3.38 m ^ j ^ 12.01 m i 14.94 ^ j
4.00
^ 2.08 m i
4.00
1.63: Assume each person sees the dentist twice a year for checkups, for 2 hours. Assume 2 more hours for restorative work. Assuming most dentists work less than 2000 hours per year, this gives 2000 hours 4 hours per patient 500 patients per dentist.
1.58: a) b) c)
($4,950 ,000 102 acres) (1acre 43560 ft 2 ) ($12 m 2 ) (2.54 cm in) 2 (1 m 100 cm) 2 $. 008 in 2 (1in 7 8 in )
10 .77 ft 2 m 2 $. 008 in 2 .
$12 m 2 .
$. 007 for postage stamp sized parcel.
1.59: a) To three significant figures, the time for one cycle is
1 1.420 109 Hz 7.04 10
3600 s 1h
10
s.
cycles h
b)
1.420 10 9
cycles s
5.11 10 12
c) Using the conversion from years to seconds given in Appendix F, 3.156 107 s 9 1.42 10 Hz 4.600
1.65: Let D be the fourth force.
A B C Ax Bx
D
0, so D
A B C A cos30.0 B cos30.0 50.00 N 69.28 N 31.90 N
A cos30.0 B sin 30.0
86.6 N, Ay 40.00 N, B y
Cx C cos53.0 24.07 N, C y C sin 53.0 22 .53 N, Dy 87 .34 N Then D x
D
2 Dx 2 D
1.54: a) From Eq. (1.22), the magnitude of the cross product is
12 .0 m 18 .0 m sin 180 37 130 m 2
The right-hand rule gives the direction as being into the page, or the z-direction. Using Eq. (1.27), the only non-vanishing component of the cross
1.55: In Eq. (1.27), the only non-vanishing component of the cross product is
so A B Cz Ax By Ay Bx 4.00 2.00 3.00 5.00 23 .00 ,
^ 23 .00 k , and the magnitude of the vector product is 23.00.
Newtons Second Law Lab Report
Chumei, Rui
02/12/2017
PHYS 181
TA: Schott Michael Louis
Lab Partner: Holly Nicole & Hardy, Patrick Jordan
Abstract
The experiment is to find the relationship between mass, force, and acceleration in formula
which is Newtons