## PHY 1150

## First Hour Exam

## February 16, 1998

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This is a multiple-choice exam. That means there is no partial credit. But it also means there are no multiple-part questions. Take your time. None of the questions are intended to be "trick questions". I think I have used

g = 10 m/sthroughout. If you do a calculation and have something like 12.345 for an answer and I have 12.346, it is probably just a round-off error (mine or yours).^{2}All I want is the "scantron" sheet, the "bubble sheet", or the answer sheet. You can keep the questions from the exam. The solutions to the questions will be posted on the world wide web tomorrow. The results of the exam will be posted, by social security number, outside my door (Room 126) shortly after they are back from Academic Testing Services.

For every question, also consider, as a possible answer,

E)none of the above.

Possibly useful information(if you think of other equations that might be useful, come see me and I'll put them on the board for everyone):s = s

_{i }+ v_{i }t + (^{1}/_{2}) a t^{2}v = v

_{i }+ a tv

^{2}= v_{i }^{2}+ 2 a ( s - s_{i})sin = opp/hyp

cos = adj/hyp

tan = opp/adj

F= ma

F_{12}= -F_{21}

V= V_{x}i+ V_{y}j+ V_{z}kw = mg

g = 10 m/s

^{2}F

_{f,max}= µ nv = r

a

_{c}= v^{2}/ rF

_{c }= m v^{2}/ rF

_{g}= G M m / r^{2}G = 6.67 x 10

^{-11}N m^{2}/ kg^{2}M

_{E}= 5.98 x 10^{24}kgR

_{E}= 4 200 km = 4.2 x 10^{6}mFor every question, also consider, as a possible answer,

E)none of the above.

1. The speed of sound in air is about 330 m/s. You observe a lightning bolt strike a tree 1.5 km away. How much time will elapse between your seeing the lightning bolt and hearing the thunder that accompanies it?A) 1.7 s

B) 4.5 sC) 7.5 s

D) 9.8 s

v = x/t

x = vt

t = x/v = (1 500 m) / (330 m/s)

t = 4.5 s

2.A car accelerates from rest to 90 km/h or 25 m/s in 8.8 s. What is its average acceleration in m/s2 ?A) 2.0 m / s

^{2}B) 4.9 m / s

^{2}C) 7.5 m / s

^{2}D) 9.8 m / s

^{2}

E) None of the abovev = v

_{i}+ atv = a t

a = v /t

v = 90 km/h [ 1 000 m / km ] [ h / 3 600 s ] = 25 m / s

a = [ 25 m / s ] / 8.8 s

a = 2.84 m / s^{2}

3.A jet aircraft landing on an aircraft carrier is brought to a complete stop from a velocity of 215 km/h in 250 m. What is its average acceleration in m/s^{2}?A) - 2.9 m/s

^{2}B) - 3.8 m/s

^{2}

C) - 7.1 m/s^{2}D) - 13.8 m/s

^{2}v

_{i }= 215 km/h [ 1 000 m / km ] [ h / 3 600 s ] = 59.7 m / sv

^{2}= v_{i }^{2}+ 2 a (s - s_{i})0 = (59.7 m / s)

^{2}+ 2 a (250 m)a = - (59.7 m / s)

^{2}/ 500 m = - 7.1 m/s^{2}

a = - 7.1 m/s^{2}

4.A certain car has an acceleration of 2.5 m/s^{2}. Assume that its acceleration remains constant. Starting from rest, how far does the car travel to reach a velocity of 90 km/h or 25 m/s?A) 12.5 m

B) 25 m

C) 75 m

D) 125 mv = 90 km/h [ 1 000 m / km ] [ h / 3 600 s ] = 25 m / s

v

^{2}= v_{i}^{2}+ 2 a (s - s_{i})(25 m/s)

^{2}= 0 + 2 (2.5 m/s^{2}) (s - s_{i})625 m

^{2}/s^{2}= (5 m/s^{2}) (s - s_{i})s - s

_{i}= (625 / 5 ) m

s - s_{i}= 125 m

5. What is the resultant -- the vector sum -- of the following vectors;That is, find

R=A+Bfor

A= 10 km, 60° W of N,B= 25 km, 37° E of SA)

R= 12.7 km, 23° E of NB)

R= 15 km, 28° E of S

C) R = 16.3 km, 23° E of SD)

R= 35 km, 37° W of N

R=A+Breally meansR_{x}= A_{x}+ B_{x}and R_{y}= A_{y}+ B_{y}R_{x}= - 8.66 km + 15 km R_{y}= 5.00 km - 20 km R_{x}= 6.34 km R_{y}= - 15 kmNow, we need to recombine these components:

R = SQRT[ (R

_{x})^{2}+ (R_{y})^{2}]R = [ (6.34)

^{2}+ (-15)^{2}] km

R = 16.3 kmtan = opp/adj = R

_{x}/ R_{y}= 6.34/15 = 0.423

= 23°

6.A book is thrown horizontally out a residence hall window 5.00 m above the grass outside the residence hall. The book lands on the grass 8.00 m from the base of the hall. Find the velocity with which the book was thrown. Use g = 10 m/s^{2}.A) 4 m/s

B) 6 m/s

C) 8 m/sD) 10 m/s

How long is the book in the air?

I have used

downaspositive, making a = g = 10 m/s^{2}y = (

^{1}/_{2}) a t^{2}t

^{2}= 2 y / a = 2 ( 5.0 m ) / ( 10 m/s^{2})t

^{2}= 1 s^{2}t = 1 s.

During that time, the book travels horizontally x = 8.00 m

x = v

_{x}tv

_{x}= x / t

v_{x}= 8 m/s

7. A block starts from rest and slides down a frictionless inclined plane that is inclined 20° from the horizontal and is 2.53 meters long. What is the acceleration of the block?A) 2.53 m/s

^{2}

B) 3.42 m/s^{2}C) 7.32 m/s

^{2}D) 9.40 m/s

^{2}F

_{net}= m aF

_{net}= mg sin 20° = (m g) (0.342)m a = (m g) (0.342)

a = 0.342 g

a = 3.42 m/s^{2}

8. A 50 ton (50,000 kg) rocket is acted on by an upward thrust of 600 kN. What is the initial acceleration of the rocket as it just lifts off the pad?A) 1.2 m/s

^{2}

B) 2.0 m/s^{2}C) 3.0 m/s

^{2}D) 12.0 m/s

^{2}F

_{net}= Thrust - WeightF

_{net}= 600 kN - M gF

_{net}= 6 x 10^{5}N - (500 000 N)F

_{net}= 6 x 10^{5}N - 5 x 10^{5}N = 1 x 10^{5}5 NF

_{net}= M a[5 x 10

^{4}kg] a = 10 x 10^{4}N

a = 2 m/s^{2}

9. A 7-kg package sits on a rough but level floor. The coefficient of sliding friction between the package and the floor is 0.27. A horizontal cord is attached to the package . What must be the tension in the cord to move the package with an acceleration of 1.0 m/s^{2}?A) 9.8 N

B) 11.9 N

C) 17.3 N

D) 25.9 Nn = m g (not always! but it is true this time).

n = m g = (7 kg) ( 10 m/s2 ) = 70 N

F

_{f}= µ n = (0.27) (70 N) = 18.9 NF

_{net}= T - F_{f}= ( 7 kg ) ( 1.0 m/s^{2}) = m aT - 18.9 N = 7.0 N

T = 18.9 N + 7.0 N

T = 25.9 N

10.A horse moves with a tangential speed of 1.9 m/s when it is 8.5 m from the center of a carousel. Calculate its centripetal acceleration.

A) 0.42 m/s^{2}B) 1.21 m/s

^{2}C) 5.2 m/s

^{2}D) 30.7 m/s

^{2}a

_{c}= v^{2}/ ra

_{c}= (1.9 m/s)^{2}/ 8.5 m

a_{c}= 0.42 m/s^{2}

11.A coin sits 0.15 m from the center of a variable-speed turntable. The coin remains in place as the speed of the turntable increases until it reaches a rate of 60 revolutions per minute and then it starts to slide. What is the coefficient of friction between the coin and the surface?A) 0.23

B) 0.37

C) 0.59D) 0.63

v = r

v = (0.15 m) ( 60 rev/min) [ 2 ¹ / rev ] [ min/60 s] = 0.94 m/s

a

_{c }= v^{2}/ ra

_{c }= (0.94 m/s)^{2}/ 0.15 m = 5.89 m/s^{2}F

_{net }= F_{f}= µ nn = m g (This is not always true; but it is true this time!)

F

_{net}= F_{c}= m a_{c}µ n = µ m g = m a

_{c}µ = a

_{c}/ gµ = 5.89 / 10

µ = 0.589

12.A car of mass 1000 kg goes over the crest of a hill whose radius of curvature is 40 m measured in a vertical plane. What is the force of the car on the road surface if the car's speed is 15 m/s?A) 3 2 00 N

B) 4 400 NC) 5 600 N

D) 10 000 N

F

_{net }= m g - n = m v^{2}/r = F_{c}n = m g - m v

^{2}/rn = m [ g - v

^{2}/ r ]a

_{c}= v^{2}/ r = (15 m/s)^{2}/ 40 m = (225/40) m/s^{2}= 5.6 m/s^{2}n = (1 000 kg) [ (10 - 5.6) m/s

^{2}]n = (1 000 kg) ( 4.4 m/s2 )

n = 4 400 N

13. What orbital radius should a weather satellite have if it is to have a period of 6.0 hours?

A) 16 800 kmB) 18 600 km

C) 26 300 km

D) 42 000 km

T = 6 h [ 3600 s / h ] = 21 600 s

v = C / T = 2 ¹ r / T

F

_{net }= m v^{2}/ r = G M m / r^{2}= F_{g}v

^{2}/r = G M / r^{2}[2 ¹ r / T]

^{2}/r = G M / r^{2}[4 ¹

^{2}r^{2 }/ T^{2}] / r = G M / r^{2}4 ¹

^{2}r / T^{2}= G M / r^{2}r

^{3}= G M T^{2}/ 4 ¹^{2}r

^{3}= [ 6.67 x 10^{&endash;11}N m^{2}/ kg^{2}] [5.98 x 10^{24}kg] [2.16 x 10^{4}s]^{2}/ 4 ¹^{2}r

^{3}= 4.71 x 10^{21}mr = 1.68 x 10

^{7}mr = 1.68 x 10

^{4}km

r = 16 800 km

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