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vesna_86 [32]
3 years ago
8

A 1100 kg car is traveling around a flat 82.3 m radius curve. The coefficient of static friction between the car tires and the r

oad is .521. What is the maximum speed in m/s at which the car can take the curve?
Physics
2 answers:
Novay_Z [31]3 years ago
8 0

Answer:

The maximum speed of car will be 20.5m/sec

Explanation:

We have given mass of car = 1100 kg

Radius of curve = 82.3 m

Static friction \mu _s=0.521

We have to find the maximum speed of car

We know that at maximum speed centripetal force will be equal to frictional force m\frac{v^2}{r}=\mu _srg

v=\sqrt{\mu _srg}=\sqrt{0.521\times 82.3\times 9.8}=20.5m/sec

So the maximum speed of car will be 20.5m/sec

pantera1 [17]3 years ago
3 0

Answer:20.51 m/s

Explanation:

Given

Mass of car(m)=1100 kg

radius of curve =82.3 m

coefficient of static friction(\mu)=0.521

here centripetal force is provided by Friction Force

F_c(centripetal\ force)=\frac{mv^2}{r}

Friction Force=\mu N

where N=Normal reaction

\frac{mv^2}{r}=\mu N

\frac{1100\times v^2}{82.3}=0.521\times 1100\times 9.81

v^2=0.521\times 9.81\times 82.3

v=\sqrt{420.63}=20.51 m/s

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Complete Question

The complete question is shown on the uploaded image

Answer:

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Explanation:

From the question we are told that

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Looking at the first diagram

           At  600 MPa of stress

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            \frac{\sigma_{\Delta l} -  \sigma_{0.0015} }{ \sigma _ {0.3} - \sigma_{0.0015} } = \frac{e_{\Delta l }  - e_{0.0015}}{e_{0.3} - e_{ 0.0015}}

Substituting values

              \frac{\sigma _{\Delta l} - 450}{600 - 450} = \frac{0.1 -0.0015}{0.3 - 0.0015}

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Generally the force on each head is mathematically represented as

              F = \sigma_{\Delta l} * A

Substituting values

             F = 499.50*10^{6} * 2.8*10^{-6}

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              T = 6 * F

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Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

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d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

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We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

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