1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Tpy6a [65]
3 years ago
5

An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e

levation of 5183 ft in Denver to the highest elevation on Trail Ridge Road in the Rocky Mountains. What is the elevation at the high point of the road, in ft?
Engineering
1 answer:
Mila [183]3 years ago
3 0

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

You might be interested in
A fluid of density 900 kg/m3 passes through a converging section of an upstream diameter of 50 mm and a downstream diameter of 2
NISA [10]

Answer:

Q= 4.6 × 10⁻³ m³/s

actual velocity will be equal to 8.39 m/s

Explanation:

density of fluid = 900 kg/m³

d₁ = 0.025 m

d₂ = 0.05 m

Δ P = -40 k N/m²

C v = 0.89

using energy equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2

under ideal condition v₁² = 0

v₂² = 88.88

v₂ = 9.43 m/s

hence discharge at downstream will be

Q = Av

Q = \dfrac{\pi}{4}d_1^2 \times v

Q = \dfrac{\pi}{4}0.025^2 \times 9.43

Q= 4.6 × 10⁻³ m³/s

we know that

C_v =\dfrac{actual\ velocity}{theoretical\ velocity }\\0.89 =\dfrac{actual\ velocity}{9.43}\\actual\ velocity = 8.39m/s

hence , actual velocity will be equal to 8.39 m/s

6 0
3 years ago
Which of the following has led to a safer and more prosperous global community within the last century? the Bronze Age composite
sattari [20]

Answer:

The bronze age

4 0
3 years ago
Read 2 more answers
Hareem and Shahad are on a road trip to Florida. They pull over to get gas, and as you may know it is sold by the gallon in the
Alona [7]

Answer:

53.3

Explanation:

4×14=56

56÷1.05=53.3

4 0
2 years ago
Question 2/5
adelina 88 [10]
All of the above. Answer.
7 0
3 years ago
Two uniformly charged conducting plates are parallel to each other. They each have area A. Plate #1 has a positive charge Q whil
Karo-lina-s [1.5K]

Answer:

E = \frac{3Q}{2A\epsilon_0}

Explanation:

By Gauss Law for electric field:

E = \frac{\sigma}{2\epsilon_0}

Where \sigma is the charge density Q/A. Since we have 2 parallel  plates with different charge, the electric field at point P in the gap would be the sum of 2 field

E = E_1 + E_2

E = \frac{Q}{2A\epsilon_0} + \frac{2Q}{2A\epsilon_0}

E = \frac{3Q}{2A\epsilon_0}

5 0
3 years ago
Other questions:
  • Hey, I have a question, I was thinking that if you have engineering skills or drawing skill you could help me to start a project
    15·1 answer
  • Science, Technology, Engineering & Mathematics
    14·1 answer
  • WHEN WAS THE FIRST CAR INVENTED?
    13·2 answers
  • If the head loss in a 30 m of length of a 75-mm-diameter pipe is 7.6 m for a given flow rate of water, what is the total drag fo
    13·1 answer
  • Which of the following tells the computer wha to do
    12·1 answer
  • Match the scenario to the problem-solving step it represents.
    7·1 answer
  • PLEASE HELPPPPPPP!!!!,
    10·2 answers
  • Transcript
    7·1 answer
  • What is considered the greatest engineering achievement of the 20th century?
    10·1 answer
  • What are some common work contexts for Licensing Examiners and Inspectors? Select four options.
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!