Ch 5 Whal accewralion of a 304t N ea down a Question 7 caloutate the magnitude o

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Ch 5 Whal accewralion of a 304t N ea down a Question 7 caloutate the magnitude of acceleration of a snowboarder gong up a 18 slope assuming the coeffoert of friesen Nor waerd wood on wer show is 0 Calculate the acocleration of a skier heading down a 120 slope. assuming he coeroent of tidion for wared wood Question 9 When rebuidng her car's engine, a physics major must evert 334 Not force to insert a dry steel pistoninto a steel oylinder What is he magnilude of normal force between the piston and Dylander he coeffcient of kinastic fricton is o.24 Question 10 structures on Earth. A 69 kg physicist placed himself and 431 kg equipment at the top at re the tallest attidal perform gravity experiments, By how much steer cylinder 0,13 m in radius? Young's modulus for steel is 2. x 10" Nmr Give your answerin mameters Question 11 What is ee lor exerted on the arip by a 1014 kg artilery shel ned with an acceleration of 27 km/sh Give arawer maona al Newunli L Question 12 Star al hotel pushes kg oart through the halls of the hotel, The caM has a el of64 TN

Explanation / Answer

6)Weight of car = 3041 N

Mass of car = 3041/9.8 = 310.3 kg

Slope of incline =12o

Coefficient of kinetic friction = n=0.12

net force acting on the car = mgsin12o-nmgcos12o = 3041xsin12o - 0.12x3041cos12o = 275.3 N

acceleration = 275.3/310.3 = 0.887 m/s2 downwards along the slope

(Note : acceleration is basically equal to gsin12o-ngcos12o and is independent of mass or weight of car)

7)Since the body is moving up the slope, the friction force is also acting downwards along the slope

Coeffieicient of friction =n= 0.11

Total acceleration = gsin18o + ngcos18o = 4.09m/s2 downwards along the slope

8)This is similar to question 6

coffiicient of fricition = n=0.11

Acceleration = gsin12o-ngcos12o =0.1 m/s2 downwards along the slope

9)Coefficient of kinetic friction = n=0.24

Minimum force needed to insert piston into cylinder = 334Newton

Let normal force be Nf

nNf = 334

Nf = 334/0.24 = 1391.67 Newtons

10)Total compressive force on antenna = (60+431)x9.8 = 4811.8N

Area of cross section of antenna = pix0.132 =0.053m2

Compressive stress = 4811.8/0.053 = 90788.7 N/m2

Young's Modulus = 2.1x1011 N/m2

By Hooke's Law, Compressive strain = 90788.7/2.1x1011 = 4.32x10-7

height of antenna = 544m

Compression in antenna = 544* 4.32x10-7 = 2.35x10-4 m = 0.235mm

11) Mass of ship = 1014kg

Acceleration = 27km/s2 = 27000m/s2

Per newton's second law, F =ma = 1014*27000 = 27378000N = 27.378 million Newtons

12) Force = 64.7N

mass = 5.91kg

Per Newton's Second Law, F=ma

a = F/m = 64.7/5.91 = 10.95m/s2

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