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

8 pages77 viewsFall 2018

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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 ×10−27 kg neutron mass 1.67 ×10−27 kg

electron mass 9.11 ×10−31 kg electron charge 1.6×10−19 C

speed of light 3.0×108m·s−1gravity acceleration 9.8 m·s−2

derivative of a power d

dx (xn) = nxn−1integral 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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