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satela [25.4K]
3 years ago
11

013 Indianapolis 500 champion Tony Kanaan holds his hand out of his IndyCar while driving through still air with standard atmosp

heric conditions. (a) For safety, the pit lane speed limit is 60 mph. At this speed, what is the maximum pressure on his hand? (b) Back on the race track, what is the maximum pressure when he is driving his IndyCar at 225 mph? (c) On the straightaways, the IndyCar reaches speeds in excess of 235 mph. For this speed, is your solution method for parts (a and (b) reasonable? Explain.
Physics
1 answer:
Stolb23 [73]3 years ago
3 0

Answer:

a) 14.76 psi

b) 15.59 psi

c) 15.68 psi

d) The solution method is reasonable

Explanation:

The formula for maximum pressure is given as using Bernoulli's Equation :

Pmax = p + 1/2ρV²

Where ρ = Density = 0.002376slug per cubic foot

p = Atmospheric pressure in pounds per cubic foot = 14.6959 pounds per square inch

Converting pounds per square inch to pounds per square foot

= 1 pounds per square inch = 144 pounds per square foot

14.6959 pounds per square inch =

14.6959 pounds per square inch × 144 pounds per square foot

= 2116.2096 pounds per square foot

V = Velocity of the flow = Speed limit.

(a) For safety, the pit lane speed limit is 60 mph. At this speed, what is the maximum pressure on his hand?

Our speed limit is 60 mph

We convert to ft/s

1 mph = 1.46667 ft/s

60 mph =

60 mph × 1.46667 ft/s = 88 ft/s

Pmax = p + 1/2ρV²

Pmax = 2116.2096 + (0.5 × 0.002376 × 88²)

Pmax = = 2125.409472 Ib/ft² =

Converting to psi.

= 2125.409472/144

= 14.759788 psi

≈ 14.76 psi

(b) Back on the race track, what is the maximum pressure when he is driving his IndyCar at 225 mph?

Our speed limit is 225 mph

We convert to ft/s

1 mph = 1.46667 ft/s

225 mph =

225 mph × 1.46667 ft/s = 330 ft/s

Pmax = p + 1/2ρV²

Pmax = 2116.2096 + (0.5 × 0.002376 × 330²)

Pmax = 2245.5828 Ib/ft² =

Converting to psi.

= 2245.5828/144

= 15.594325 psi

≈ 15.59 psi

(c) On the straightaways, the IndyCar reaches speeds in excess of 235 mph.

Our speed limit is 235 mph

We convert to ft/s

1 mph = 1.46667 ft/s

235 mph =

235 mph × 1.46667 ft/s = 344.667 ft/s

Pmax = p + 1/2ρV²

Pmax = 2116.2096 + (0.5 × 0.002376 × 344.667²)

Pmax = = 2257.338465 Ib/ft² =

Converting to psi.

= 2257.3384654/144

= 15.675961562 psi

≈ 15.68 psi

d) For this speed, is your solution method for parts (a and (b) reasonable? Explain.

It is important to note that: the value of 1 atmosphere = 14.696 psi

If we look at the solution above, we can see that the maximum pressure of the speeds in question a, b, c which are: 14.76 psi, 15.59 psi, 15.68 psi respectfully are just a little bit outside the range of 1 atmosphere.

Hence, this is not significant enough to cause damage or harm the hands of Tony Kanaan therefore, my solution method for parts (a and (b) reasonable is reasonable.

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Attempt 2 You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 15151515 l
Natasha_Volkova [10]

Answer:

v = 28.98 ft / s

Explanation:

For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision

In the exercise they indicate the weight of each car

          Wₐ = 1500 lb

          W_b = 1125 lb

Car B's velocity from v_b = 42.0 mph westward, car A travels east

let's find the mass of the vehicles

             W = mg

             m = W / g

             mₐ = Wₐ / g

             m_b = W_b / g

             mₐ = 1500/32 = 46.875 slug

             m_b = 125/32 = 35,156 slug

Let's reduce to the english system

             v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s

We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved

we assume the direction to the east (right) positive

initial instant. Before the crash

           p₀ = mₐ v₀ₐ - m_b v_{ob}

final instant. Right after the crash

           p_f = (mₐ + m_b) v

the moment is preserved

           p₀ = p_f

           mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v

           v = \frac{ m_a \ v_{oa} - m_b \ v_{ob}  }{ m_a +m_b}

we substitute the values

           v = \frac{ 46.875}{82.03} \ v_{oa} -  \frac{35.156}{82.03} \ 61.6

           v = 0.559 v₀ₐ - 26.40                  (1)

Now as the two vehicles united we can use the relationship between work and kinetic energy

the total mass is

              M = mₐ + m_b

              M = 46,875 + 35,156 = 82,031 slug

starting point. Jsto after the crash

              K₀ = ½ M v²

final point. When they stop

             K_f = 0

The work is

             W = - fr x

the negative sign is because the friction forces are always opposite to the displacement

Let's write Newton's second law

Axis y

           N-W = 0

           N = W

the friction force has the expression

            fr = μ N

we substitute

            -μ W x = Kf - Ko

             

            -μ W x = 0 - ½ (W / g) v²

            v² = 2 μ g x  

            v = \sqrt{ 2 \ 0.750 \ 32 \ 17.5}Ra (2 0.750 32 17.5  

            v = 28.98 ft / s

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