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dexar [7]
3 years ago
8

An automobile moves on a level horizontal road in a circle of radius 30 m. The coefficient of

Physics
1 answer:
blondinia [14]3 years ago
5 0

Answer:

v = 12.12 m/s      

Explanation:

It is given that,

Radius of circle, r = 30 m

The coefficient friction between tires and road is 0.5,

The centripetal force is balanced by the force of friction such that,

v = 12.12 m/s

So, the maximum speed with which this car can round this curve is 12.12 m/s. Hence, this is the required solution.

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For a body to be in SHM, the restoring force must be __________ to displacement from equilibrium.
Ipatiy [6.2K]

Answer:

Proportional

Explanation:

The conditions that must be met to produce SHM are;

-The restoring force must be proportional to the displacement and act opposite to the direction of motion with no drag forces or friction.

- The frequency of oscillation does not depend on the amplitude.

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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
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