d book, closed notes exam. You may use a scientifie caleulator. No phone caleula

Question

d book, closed notes exam. You may use a scientifie caleulator. No phone caleulators or devices may be used. Show all your work on the papers provided, not on the er sheet. ALL WORK MUST BE SHOWN CLEARLY, Return all papers to the ructor. enh-20 N,01-35°, F2=15.0 N, 2 = 120 Find the resultant force FR-Fi+F2 1019] C) 19i-13j D) 5.7i+2.3j E)-12i+14j F) None of these 8.9 i+24.5j ) Find the resultant force FR A) 26 N -70° B) 22 N =62° C) 31 N0-90 -Fi+ F D) 15N 0-120 E)42 N =15° F) None of these 3) What is the sum of 2.673 +1.976+2.100? G) None of these A) 6.7 B) 6.75 C) 6.749 D) 6.7490 E) 6.7 F) 6.8 4) A helicopter is ascending vertically with a speed of 8.20 m/s. At a height of 155 m above ground a package is dropped from a window. How much time is required for the package to hit the ground? A) 6.18 s B) 5.61 s C) 9.55 s D) 8.34 s E) 5.57 F) None of these 5) An object is thrown upward with a speed of 12 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/s. How long does it take for the object to reach the maximum height? A) 8.0s B) 3.1 s E)4.8 s F) None of these D) 16 s 6) At what angle should a water-gun be aimed in order for the water to land with the greatest horizontal range? A) 0° B) 30° C) 45° D) 60° E) 15° F) None of these ) A girl throws a rock horizontally, with a velocity of 10 m/s, from a bridge. It falls 20 m to the ater below. How far does the rock travel horizontally before striking the water? )14 m 16 m C) 20 m D) 24 m E) 27 m F) None of these

Explanation / Answer

F1 = 20*cos35i + 20*sin35 j

F2 = 15*cos120i + 15*sin120 j

FR = F1 + F2

FR = (20*cos35 + 15*cos120) i + (20*sin35 + 15*sin120)j

FR = 8.9 i + 24.5 j   <<<<<================ANSWER

==================================

2)

magnitude = sqrt(8.9^2+24.5^2) = 26 N

theta = tan^-1(24.5/8.9) = 70

OPTION ( A ) <<<<<=======ANSWER

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3)

the sum should have 3 decimal places

6.749

OPTION ( C)

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4)

initial velocity voy = + 8.2 m/s

displacement y = -155 m

acceleration ay = -g = -9.8 m/s^2

from equation of motion

y = voy*t + (1/2)*ay*t^2

-155 = (8.2*t) - (1/2)*9.8*t^2

t = 6.18 s

============================

4)

initial speed vi = 12 m/s

final speed at maximum height vf = 0

acceleration a = -2.5 m/s^2

vf = vi + a*t

0 = 12 - (2.5*t)

t = 4.8 s

OPTION ( E )

==========================

6)

range R = v^2(sin2theta)/g

for R to be maximum

sin(2theta) is maximum

maximum value of sin = 1

sin(2theta) = 1

sin(2theta) = sin90

2theta = 90

theta = 45

OPTION ( C )

============================

along vertical

initial velocity voy = 0

acceleration ay = -g

displacement y = -20 m

y = voy*t + (1/2)*a*t^2

-20 = 0 - (1/2)*10*t^2

t = 2 s

along horizontal

x = vox*t

x = 10*2 = 20 m <<<-------------ANSWER

OPTION ( C )

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