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Yuri [45]
2 years ago
10

A coffee cup on the horizontal dashboard of a car slides forward when the driver decelerates from 45 kmh to rest in 3.5 s or les

s, but not if she decelerates in a longer time. What is the coefficient of static friction between the cup and the dash? Assume the road and the dashboard is level (horizontal).
Physics
1 answer:
Svetach [21]2 years ago
6 0

Answer:

μs = 0.36

Explanation:

Assuming no other forces acting on the cup while the car is decelerating, the friction force is responsible for any horizontal movement of the cup.

If the cup is on the verge of starting to slide, the friction force can be expressed as follows:

Ff = -μs*N = -μs*m*g

This force produces a deceleration from 45 Kmh to rest, in 3.5 s or  more.

Converting 45 kmh to m/s, we have:

45 kmh *\frac{1000m}{1km} * \frac{1h}{3600 s} = 12.5 m/s

We can find the acceleration, just applying the definition, with vf =0, as follows:

a = \frac{-vf}{t} =\frac{-12.5m/s}{3.5s} = -3.57 m/s2

According to Newton's 2nd law, we can write the following expression:

F = m*a = -μs*m*g

Simplifying common terms, we can solve for μs, as follows:

μs = \frac{a}{-g} =\frac{-3.57 m/s2}{-9.80m/s2} = 0.36

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Answer:

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Explanation:

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            P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

This case does not indicate at the surface pressure is P₂, the pressure at the outlet is P₁ = P₀, the surface velocity is zero v₂ = 0

          P₀ + ½ ρ v₁² + ρ g y₁ = P₂ + 0 + ρ g y₂

           ½ ρ v₁² = P₂-P₀ + ρ (y₂ -y₁)

          v₁² = 2 (P₂-P₀) /ρ + 2 (y₂ -y₁)

          v₁ = √ [2 (P₂-P₀) /ρ + 2 (y₂ -y₁)]

b) Reduce the pressure to SI units

         P₂ = 0.86 atm (1.01 10⁵ Pa / 1 atm) = 0.8686 10⁵ Pa

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         ρ = 1 10³ kg / m³

Let's calculate

         v₁ = √ [2 (0.8686 -1.01) 10⁵/10³ + 2 (2.6)]

         v₁ = √ [-0.2828 10² + 5.2] = Ra [-23]

Water does not flow, this is because the pressure of the inner part is less than atmospheric, so that the water flows the pressure P₂> = P₀

For example if the pressure P₂ = P₀

         v₁ = √ 5.2

          v₁ = 2.28 m / s

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Answer:

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At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setti
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Answer:

Part(a): The angular acceleration is 5.63~rad~s^{-2}.

Part(b): The angular displacement is 2629~rad.

Explanation:

Part(a):

If \omega_{1},~\omega_{2}~and~\alpha be the initial angular speed, final angular speed and angular acceleration  of the centrifuge respectively, then from rotational kinematic equation, we can write

\alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)

where 't' is the time taken by the centrifuge to increase its angular speed.

Given, \omega_{i} = 250~rad~s^{-1}, \omega_{f} = 750~rad~s^{-1} and t = 9.5~s. From equation (I), the angular acceleration is given by

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Part(b):

Also the angular displacement (\Delta \theta) can be written as

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8 0
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Answer:

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