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weqwewe [10]
2 years ago
13

Alberta is going to have dinner at her grandmother's house, but she is running a bit behind schedule. As she gets onto the highw

ay, she knows that she must exit the highway within 55
min

min

if she is not going to arrive late. Her exit is 43

mi

miHow much time would it take at the posted 60 mph speed?

away.
Physics
1 answer:
viktelen [127]2 years ago
5 0

Complete Question

Alberta is going to have dinner at her grandmother's house, but she is running a bit behind schedule. As she gets onto the highway, she knows that she must exit the highway within 55 min if she is not going to arrive late. Her exit is 43 mi away. How much time would it take at the posted 60 mph speed?

Answer:

The time it would take at the given speed is  x = 43.00 \ minutes

Explanation:

From the question we are told that

      The time taken to exist the highway is  t = 55 min

      The distance to the exist is  d =  43\  mi

       Alberta speed is v =  60 mph

The time it would take travelling at the given speed is mathematically represented as

        t_z = \frac{d}{v}

  substituting values

        t_z = \frac{43}{60}

        t_z = 0.71667\ hrs

Converting to minutes

         1  hour =  60 minutes

So      0.71667 hours = x minutes

   Therefore

               x = 0.71667 * 60

                x = 43.00 \ minutes

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Pikes Peak near Denver, Colorado, has an elevation of 14,110 ft. Calculate the pressure at this elevation using three different
kramer

Answer:

a) P = 1240 lb/ft^2

b) P = 1040 lb/ft^2

c) P = 1270 lb/ft^2

Explanation:

Given:

- P_a = 2216.2 lb/ft^2

- β = 0.00357 R/ft

- g = 32.174 ft/s^2

- T_a = 518.7 R

- R = 1716 ft-lb / slug-R

- γ = 0.07647 lb/ft^3

- h = 14,110 ft

Find:

(a) Determine the pressure at this elevation using the standard atmosphere equation.

(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.

(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.

Solution:

- The standard atmospheric equation is expressed as:

                           P = P_a* ( 1 - βh/T_a)^(g / R*β)

                          (g / R*β) = 32.174 / 1716*0.0035 = 5.252

                            P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252

                            P = 1240 lb/ft^2

- The air density method which is expressed as:

                            P = P_a - γ*h

                            P = 2116.2 - 0.07647*14,110

                            P = 1040 lb/ft^2

- Using constant temperature ideal gas approximation:

                            P = P_a* e^ ( -g*h / R*T_a )

                            P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )

                            P = 1270 lb/ft^2

6 0
2 years ago
Read 2 more answers
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