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Nookie1986 [14]
3 years ago
9

Consult Multiple Concept Example 10 in preparation for this problem. Traveling at a speed of 15.4 m/s, the driver of an automobi

le suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.680. What is the speed of the automobile after 1.11 s have elapsed? Ignore the effects of air resistance.
Physics
1 answer:
ella [17]3 years ago
7 0

Answer: 8.07 m/s

Explanation:

Applying the definition of acceleration, as the rate of change of velocity, we can extract the value of the velocity v at a given time t, as follows:

v = v₀ + a t (1)

As we already know v₀ and t, our  single unknown is the acceleration a.

Once the friver slammed on the brakes, there exists an external force, due to friction, which is the only external force acting in the horizontal direction, producing a deceleration.

This friction force, is equal to the product of  the coeficient of  kinetic friction, times the normal force.

In this case, the normal force is numerically equal to the gravity force, which we call weight.

This force opposes to the direction of the initial velocity, as it always opposes to the relative movement between both surfaces in contact.

Applying Newton's 2nd Law, we get:

Fnet = ma ⇒ Ff = m a ⇒ -μk . m . g = m.a

Solving for a :

a = -μk . g = -0.68 . 9.8 m/s² = -6.66 m/s²

Replacing in (1)

v = 15.4 m/s +(-6.66) m/s². 1.11 s = 8.07 m/s

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Learning Goal: To understand the behavior ofthe electric field at the surface of a conductor, and itsrelationship to surface cha
Ivan

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a   it is always zero

b  0

c  \eta  =  -\epsilon _o E

Explanation:ss

Here the  net charge is  on the outer surface of the conductor thus this means that the net charge inside the conductor is zero

Generally the charge density of a conductor is dependent on the charge per unit area  which implies that the charge density is dependent on the net charge  so this  means that the charge density inside the conductor is zero

 

Generally the direction of electric field this from the  positive charge to the negative charge  so from the question we can deduce  that the negative charge is located on the surface of the conductor

    So We can mathematically define the charge density on the surface of the electric field as

             ∮E \cdot dA =  \frac{-Q}{\epsilon _o}

Where E is the electric field

          dA change in unit area

           -Q is the negative charge

          \epsilon _o  is the permittivity of free space

So

          EA  =  \frac{-Q}{\epsilon _o }

           \frac{Q}{A}  =  -\epsilon _o E

          \eta  =  -\epsilon _o E

Where \eta is the charge density

   

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Answer:

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Explanation:

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