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The corners of an equilateral tringle of side length 35 cm, lie on a disk, as shown in the diagram

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The corners of an equilateral tringle of side length 35 cm, lie on a disk, as shown in the diagram. The area of the disk, in square inches, is equal to oe

 

. 35

a) 50 in? d =—— "= 13.78 in a

b) 150 in? 2.54cm/in ; .

c) 200in — df2 | | ° !

d) 600 in? r= cos30° ein 35cm

e) 800 in? A= nr2 7 -

 

2. A driver requires a 1.2 s reaction time before stepping on the brakes. While travelling at

90 km/h, the driver sees a moose crossing the road 74 m ahead. The minimum

deceleration the car must undergo to avert an accident is equal to

a) 3.0 ms? The stopping distance d is the sum of reaction and braking distances.

b) 4.2 m/s’ v2

c) Films d=v; (+50

 

2

ee yas

=> SO

0 d-—v,t, 74m-—(25m/s)(1.2s)

 

3. A hot-air balloon leaves the ground and accelerates upward at 0.25 m/s’. After 56 s, a

piece of rock loses contact with the bottom of the balloon’s basket. The maximum height

of the rock above the ground 1s equal to

 

The upward velocity and elevation of the rock, at the instant

a) 10m it detaches from the balloon’s basket. are

b) 24m 2

c) 380m Vc, = at = (0.25 m/s*)(56s) = 14m/s

d) 390m Yeo = ~at? = 392m

e) 400m 3

0 = veg — 29 (Wmax — Yse) = Ymax = 402m

4. In a training session, a cyclist starts from rest and accelerates uniformly down an inclined track. The coach records a travel distance of 60 m during the second 5.0 s of motion. The cyclist’s acceleration 1s equal to ve = a(5.0s)

a) 1.2 ms’ 60 m = vs (5.0s) +24 (5.05)?

b) 1.6ms- 2

c) 2.4 m/s? => 60m = 5a (5.0 s)?

d) 4.8 m/s?

e) 5.0 ms? 5g - 260m)

3 (5.0 s)2

5. If A+ B=3.0i1+2.0k and A-B=5.0i-4.0j + 4.0k, the vector A 1s equal to

a) i+2.0j-k _ 8 _

b) i+3.0j-2.0k (A + B) + (A-B) = 2A = 8.0i- 4.0j + 6.0k

c) -2.01+4.0j-2.0k 71 -  a0¢

dy) 401 20] 30k => A=,(4.0i-2.0j+3.0k)

e) 8.0i1-4.0j+60k

6. A car travels at 30 km/h west, makes a left turn, and emerges at 21 km/h south, as shown in the diagram. During the turn, the average acceleration of the car is

a) directed 35°. south of east. ~ ease

b) directed 35°, east of south. 0 vo 30km/h

c) directed 35°, south of west. ’ N

d) directed 35°, west of south. aD vy; 

e) equal to zero. Av W --E

AW -   21km/h 5

a=a and v,; =v, + Av 

= tan? (VL) = tan-3 (2

@ = tan (1) = tan (=)

7. A particle starts at the origin with an initial velocity of (6.0 i + 4.0 j) m/s and moves in the xy plane at a constant acceleration of (—4.0 i + 8.0 j) m/s”. Three seconds later, the particle is located at a distance from the origin equal to

a) 29m X3 = Vyit + 5a,t? = (6.0)(3.0) +> (—4.0)(3.0)? = 0

b) 40m 1

c) 43m y3 = (4.0)(3.0) + = (8.0)(3.0)? = 48 m

d) 48m 

e) 86m =>d= [x2 + y3

8. In a training camp, a target 1s dropped from a helicopter hovering at an elevation of 54 m above the ground. One second later, an archer in a tower, located 120 m away, aims in the horizontal direction and shoots an arrow at 60m/s and hits the target, as shown in the diagram. Neglect wind and air resistance. The elevation above the ground at which the arrow hits the target is equal to

a) 9.9 m
b) 20 m
c) 30 m
d) 34 m
e) 44 m
 

9. A moving sidewalk in an airport terminal transports a standing traveller in 60 s. The same person can walk an equal distance in 40 s. When walking on the moving sidewalk, the transportation time is equal to

a) 20s d=v,(70s) = %4,(55s)

by) 4s While walking on the moving sidewalk, transportation time becomes

2 ie: ___4 d (60 s) (40s)

 

S en

(uUtvy) (dad , d_ 605+ 40s

e) 350s (cos + a03)

 

10... A car starts from rest and travels around a circular track, of radius r = 144 m, at increasing speed given by v = (3.0 ¢) m/s, where ¢ 1s in seconds. The magnitude of the

car’s acceleration at ¢ = 8.0 s, is equal to

a) 1.0 m/s? Vg = 3.0 (8.0) = 24m/s ~

b) 2.0 m/s” dv 3

c) 40ms? a =a = 3.0 m/s

d) S0ms 2 oo yp —_+|

e) 7.0 m/s’ a, = = = 40m/s?

 

a= la? +a?

11. | Asystem of three blocks, on a smooth table, is pulled by a 36-N horizontal force, as

shown in the diagram. The net force on the 6.0-kg block is equal to

6.0 N 36 .N

a) y a = —— = 2.0m/s? 36 N

by) I2N 18 kg ———————

6 kg Oke

c) I8N F, = (6.0 kg) a 3 kg

d) 30N ° |

e) 36N

 

12. A box 1s projected up a ramp, inclined at 28°, with an initial speed of 10 m/s, as shown in

the diagram. If the coefficient kinetic friction between the box and the surface of the

ramp is 0.40, the box comes to a stop after travelling a distance equal to

a) 4.8m a = —gsin@ — p, gcos@ = —8.06 m/s” ;

b) 6.2m 0=v2+2ad ~

c) Ilm , \s

d) 14m __% _ _ (0m/s)* gwl

e) 44m aren 2a yw

3. A light aluminum pulley is suspended by a spring balance, as shown in the diagram. Two

blocks (mm = 6.0 kg and m2 = 2.0 kg) hang from the ends of a light rope that passes over

the pulley. The spring balance reads

a) 2.0kg a= (+) g = 4.9 m/s?

 

b) 3.0kg mye Me

c) 4.0kg T = m,(g - a) = 29.4N I,

d) 6.0 kg T, = 2T = 58.8N V4) |

e) 80k

 

) 80%K8 cor __f, flr

 

pring balance reading = 7

m2

Wty

 

14. A 70-kg woman stands on a scale in an elevator while accelerating upward at

a = 1.4 m/s’, as shown in the diagram. The scale reads

a) 60kg n—mg = ma |

b) 70kg = = 784N -

 

c) SOkg n=m(g + " a

d) 90kg Mapparent — > co {<

e) 98kg 9 aan

 

=m

 

15. | A 250-g bob 1s attached to the end ofa light string of length 25 cm. The system rotates in

a vertical circle, where the bob has a speed of 2.5 m/s at the highest point of the swing, as

shown in the diagram. The tension in the string at this position 1s equal to

a) 87IN T + mg _mv* 2.9 m/s ;

 

b) 6.2N L ae ae

c) 3.8 N mv? “mg ‘s

d) 2.4N r=—_7—mg ’ ro

€) zero. (0.25 kg)(2.5 m/s)?

= 025m © _ (0.25 kg) g ‘. i“
16. A 1.0-kg bob is attached to a vertical rod by two 1.0-m long light wires, and the points of

attachment are 1.0 m apart, as shown in the diagram. If the bob rotates in a honzontal

circle at a uniform speed of 4.1 m/s, the tension in the upper wire 1s equal to

a) 14T

b) 4.9N Im

c) 9.8N al ---- >

d) 14N CO 1 sy

e) 2EN ot eee

 

r = ¥(1.0 m)2 — (0.50 m)2 = 0.866m (radius of rotation) a m

. ° . 0 mg

T, sin 30° —T,sin30°-mg=0 = Ty — Th = = 3°

T, cos 30° +7, cos30° = eo Ty =e

r r cos 30°

By adding the equations above, the tension in the upper string ts calculated to be

1 9g v?

hh= zm (=: 30° * rcos =)