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4 years ago

12

________ is the stress caused by forces that slip or slide one part of a material against another. a) Bending b) Shear c) Compre

ssion

1 answer:

jenyasd209 [6]4 years ago

8 0

The best answer to fill in the blank would be B) Shear because shear stress is the process of where the composition of an object changes by sliding forces and so because of this, its a deformity.

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A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0

3 years ago

The gravitational potential is a distance z=0.17 m away from a distribution of mass that has the following equation:

sveta [45]

gravitational potential is given as

$V = \frac{GM}{R^2}(R^2 + Z^2)^0.5$

$E = -\frac{dV}{dz}$

$E = -\frac{GM}{R^2}\frac{d}{dz}(R^2+z^2)^0.5$

$E = -\frac{GM}{R^2}*0.5(R^2+z^2)^-0.5*2z$

$E = -\frac{GM*z}{R^2*(R^2 + z^2)^0.5}$

now plug in all values given

M = 110 kg

R = 0.55 m

z = 0.17 m

$E = -\frac{6.67 * 10^{-11}* 110*0.17}{0.55^2*(0.55^2 + 0.17^2)^0.5}$

$E = 7.16 * 10^{-9} N/kg$

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