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Midterm

PHY 131 Study Guide - Midterm Guide: Arc Welding, Dimensional Analysis, Electron Rest MassExamPremium

8 pages77 viewsFall 2018

Department
Physics
Course Code
PHY 131
Professor
All
Study Guide
Midterm

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PHY131 Midterm — Doe, Jane XXXXXXXXX (form: 1) 1
PHY131 Spring 2017, Midterm 1
Doe, Jane
XXXXXXXXX
Academic Integrity Statement
Please sign on side two of the answer sheet (over ”General Purpose Answer Sheet”). Your signature
acknowledges that your work on this exam has been done in accordance with an expectation of
academic honesty.
Fine Print
This is a closed book exam. You may use one sheet of notes - 81
2×11 inches, double sided, handwritten. ANY
manipulation of electronic equipment (iPod, iWatch, phone, laptop computer, taser, arc welder, etc.) other than
your calculator is strictly prohibited, and will result in a zero on this exam.
Your name and student identification number are printed on this exam. Before you start, enter your name and
student number on the answer sheet. Mark 1 in the “Test Version” box. Answer all problems on the bubble scan
sheet. If you think there is more than one correct answer for a given question, bubble in all of the correct answers.
Some potentially useful expressions and constants are:
proton mass 1.67 ×1027 kg neutron mass 1.67 ×1027 kg
electron mass 9.11 ×1031 kg electron charge 1.6×1019 C
speed of light 3.0×108m·s1gravity acceleration 9.8 m·s2
derivative of a power d
dx (xn) = nxn1integral of a power Rxndx =1
n+1 xn+1
Each question carries equal weight. Answer each problem with the answer closest to the correct answer. When
mistakes are found, the problematic question will be excluded from the score.
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PHY131 Midterm — Doe, Jane XXXXXXXXX (form: 1) 2
1. Have you signed side two of the answer sheet over the words ”General Purpose Answer Sheet”
to acknowledge that your work on this exam has been done in accordance with an expectation of
academic honesty?
A. No I haven’t.
B. That’s to much trouble.
C. What answer sheet?
D. I don’t want a free point.
E. Yes! I have signed side two of the answer sheet over the words ”General
Purpose Answer Sheet!”
SOLUTION: You signature on the answer sheet!
2. You’re friend is trying to convince you that there is a relationship between mass (m), acceleration
(a), velocity (v) and elapsed time (t) and a quantity they call the vibrancy (which is defined as
mass times velocity squared). Use dimensional analysis to decide which relation might be correct.
A. mavt =mv2
B. 1
2mat2=mv2
C. mvt2=mv2
D. maxt =mv2
E. None of the above.
SOLUTION: The dimensions for each value are [M] for mass, [T] for time, [L]/[T] for
velocity and [L]/[T]2for acceleration. So trying each possible answer:
mavt =mv2[M][L]
[T]2
[L]
[T][T][M][L]2
[T]2(1)
1
2mat2=mv2[M][L]
[T]2[T]26≡ [M][L]2
[T]2(2)
mvt2=mv2[M][L]
[T]2[T]26≡ [M][L]2
[T]2(3)
maxt =mv2[M][L]
[T]2[T]26≡ [M][L]2
[T]2(4)
So the solution is Equation 1.
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PHY131 Midterm — Doe, Jane XXXXXXXXX (form: 1) 3
3. You are told that the charge of an electron, e, has dimensions [Q], and that a volt, V, has
dimensions [M][L2]
[Q][T]2, the speed of light has dimensions [L]/[T], and momentum has dimensions of
[M][L]/[T]. Use dimensional analysis to determine how to express the momentum in terms of e,V,
and c.
A. eV
c
B. eV
c2
C. eV c
D. eV
E. None of the above
SOLUTION: The best way to try this is to figure out the dimensions of each case.
eV
c[Q][M][L2]
[Q][T]2
[T]
[L]=[M][L]
[T][M][L]
[T](5)
eV
c2[Q][M][L2]
[Q][T]2
[T]2
[L]2=[M][L]2
[T]26≡ [M][L]
[T](6)
eV c [Q][M][L2]
[Q][T]2
[L]
[T]=[M][L]3
[T]36≡ [M][L]
[T](7)
eV [Q][M][L2]
[Q][T]2=[M][L]2
[T]26≡ [M][L]
[T](8)
So the correct relation is Equation 5.
4. Estimate the number of cars per lane that are stuck on the LIE during rush hour. Assume that
the evening traffic coming out of the city on the LIE stretches for 20.0 km, and the average space
between cars is 25.0 m.
A. 800 cars/lane
B. 0.800 cars/lane
C. 500 cars/lane
D. 2.00 ×104cars/lane
E. 2.50 ×104cars/lane
SOLUTION: Since the distance between cars is 25.0 m, the number of cars per meter
is 1 car
25.0m. Since the traffic stretchs for 20.0 km we can find the total number of cars
20.0 km
1 lane !1000 m
1 km 1 car
25.0 m= 800 cars/lane.(9)
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