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Aliun [14]
2 years ago
13

For typical rubber-on-concrete friction, what is the shortest time in which a car could accelerate from 0 to 70 mph? suppose tha

t μs=1. 00μs=1. 00 and μk=0. 80μk=0. 80
Physics
1 answer:
algol132 years ago
3 0

The shortest time in which a car could accelerate from 0 to 70 mph is 3.193 seconds.

To find the answer, we have to know more about the friction.

<h3>How to find the shortest time?</h3>
  • We have the expression for final velocity as,

                          v=u+at

  • It is given in the question that,

                        v=70mph=31.29m/s\\1mile=1609.3m,1hour=3600s\\u=0\\k_s=1\\k_k=0.8

  • It is given that; the car could accelerate from 0 to 70mph. Thus, acceleration will be,

                    a=g*k_s=9.8*1=9.8m/s^2

  • Thus, the shortest time will be,

                    t=\frac{v-u}{a} \\t=\frac{31.29}{9.8}=3.193s

Thus, we can conclude that, the shortest time in which a car could accelerate from 0 to 70 mph is 3.193 seconds.

Learn more about the shortest time here:

brainly.com/question/3017271

#SPJ4

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Correct temperature is 80°F

Answer:

T_f = 38.83°F

Explanation:

We are given;

Volume; V = 8 ft³

Initial Pressure; P_i = 100 lbf/in² = 100 × 12² lbf/ft²

Initial temperature; T_i = 80°F = 539.67 °R

Time for outlet flow; t_o = 90 s

Mass flow rate at outlet; m'_o = 0.03 lb/s

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Now, from ideal gas equation,

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Thus;

v = RT/P

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v_i = 2 ft³/lb

Formula for initial mass is;

m_i = V/v_i

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m_i = 4 lb

Now change in mass is given as;

Δm = m'_o × t_o

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Δm = 2.7 lb

Now,

m_f = m_i - Δm

Thus; m_f = 4 - 2.7

m_f = 1.3 lb

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v_f = 8/1.3

v_f = 6.154 ft³/lb

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Pv = RT

Thus;

T_f = P_f•v_f/R

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T_f = 498.5°R

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T_f = 38.83°F

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