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
statuscvo [17]
3 years ago
8

A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric c

onditions, what is the maximum pressure on her hand? What would be the maximum pressure if the "car" were an Indy 500 racer traveling 200 mph?
Engineering
1 answer:
Paraphin [41]3 years ago
5 0

Answer:

10.8\ \text{lb/ft^2}

101.96\ \text{lb/ft}^2

Explanation:

v_1 = Velocity of car = 65 mph = 65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}

\rho = Density of air = 0.00237\ \text{slug/ft}^3

v_2=0

P_1=0

h_1=h_2

From Bernoulli's law we have

P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}

The maximum pressure on the girl's hand is 10.8\ \text{lb/ft^2}

Now v_1 = 200 mph = 200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}

P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2

The maximum pressure on the girl's hand is 101.96\ \text{lb/ft}^2

You might be interested in
Consider fully developed laminar flow in a circular pipe. If the viscosity of the fluid is reduced by half by heating while the
gladu [14]

Answer:

The pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its original value.

Explanation:

For a fully developed laminar flow in a circular pipe, the flowrate (volumetric) is given by the Hagen-Poiseulle's equation.

Q = π(ΔPR⁴/8μL)

where Q = volumetric flowrate

ΔP = Pressure drop across the pipe

μ = fluid viscosity

L = pipe length

If all the other parameters are kept constant, the pressure drop across the circular pipe is directly proportional to the viscosity of the fluid flowing in the pipe

ΔP = μ(8QL/πR⁴)

ΔP = Kμ

K = (8QL/πR⁴) = constant (for this question)

ΔP = Kμ

K = (ΔP/μ)

So, if the viscosity is halved, the new viscosity (μ₁) will be half of the original viscosity (μ).

μ₁ = (μ/2)

The new pressure drop (ΔP₁) is then

ΔP₁ = Kμ₁ = K(μ/2)

Recall,

K = (ΔP/μ)

ΔP₁ = K(μ/2) = (ΔP/μ) × (μ/2) = (ΔP/2)

Hence, the pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its value.

Hope this Helps!!!

4 0
3 years ago
Ninety-five percent of the acetone vapor in an 85 vol.% air stream is to be absorbed by countercurrent contact with pure water i
Sati [7]

Answer:

Explanation:

.......................................................................................................................

5 0
3 years ago
(Gas Mileage) Drivers are concerned with the mileage their automobiles get. One driver has kept track of several trips by record
Effectus [21]

Answer:

import java.util.*;

public class Main {

   

   public static void main(String[] args) {

     

       double milesPerGallon = 0;

       int totalMiles = 0;

       int totalGallons = 0;

       double totalMPG = 0;

       

       Scanner input = new Scanner(System.in);

 

       while(true){

           System.out.print("Enter the miles driven: ");

           int miles = input.nextInt();

           if(miles <= 0)

               break;

           else{

               System.out.print("Enter the gallons used: ");

               int gallons = input.nextInt();

               totalMiles += miles;

               totalGallons += gallons;

               milesPerGallon = (double) miles/gallons;

               totalMPG = (double) totalMiles / totalGallons;

               System.out.printf("Miles per gallon for this trip is: %.1f\n", milesPerGallon);

               System.out.printf("Total miles per gallon is: %.1f\n", totalMPG);

           }

       }

   }  

}

Explanation:

Initialize the variables

Create a while loop that iterates until the specified condition is met inside the loop

Inside the loop, ask the user to enter the miles. If the miles is less than or equal to 0, stop the loop. Otherwise, for each trip do the following: Ask the user to enter the gallons. Add the miles and gallons to totalMiles and totalGallons respectively. Calculate the milesPerGallon (divide miles by gallons). Calculate the totalMPG (divide totalMiles by totalGallons). Print the miles per gallon and total miles per gallon.

6 0
3 years ago
First person to awnser gets brainlyest!!​
Andrews [41]

Answer:

ananswer my question please

7 0
3 years ago
Read 2 more answers
How much energy does it take to boil water for pasta? For a one-pound box of pasta
ArbitrLikvidat [17]

Answer:

a.  164 °F b. 91.11 °C c. 1439.54 kJ

Explanation:

a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?

Since the starting temperature is 48°F and the final temperature which water boils is 212°F, the number of degrees Fahrenheit we would need to raise the temperature is the difference between the final temperature and the initial temperature.

So, Δ°F = 212 °F - 48 °F = 164 °F

b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?

To find the degree change in Celsius, we convert the initial and final temperature to Celsius.

°C = 5(°F - 32)/9

So, 48 °F in Celsius is

°C₁ = 5(48 - 32)/9

°C₁ = 5(16)/9

°C₁ = 80/9

°C₁ = 8.89 °C

Also, 212 °F in Celsius is

°C₂ = 5(212 - 32)/9

°C₂ = 5(180)/9

°C₂ = 5(20)

°C₂ = 100 °C

So, the number of degrees in Celsius you must raise the temperature is the temperature difference between the final and initial temperatures in Celsius.

So, Δ°C = °C₂ - °C₁ = 100 °C - 8.89 °C = 91.11 °C

c. [2 pts] How much energy is required to heat the four quarts of water from

48°F to 212°F (boiling)?

Since we require 15.8 kJ for every degree Celsius of temperature increase of the four quarts of water, that is 15.8 kJ/°C and it rises by 91.11 °C, then the amount of energy Q required is Q = amount of heat per temperature rise × temperature rise =  15.8 kJ/°C × 91.11 °C = 1439.54 kJ

8 0
3 years ago
Other questions:
  • Steam at 40 bar and 500o C enters the first-stage turbine with a volumetric flow rate of 90 m3 /min. Steam exits the turbine at
    9·1 answer
  • A turbine operates at steady state, and experiences a heat loss. 1.1 kg/s of water flows through the system. The inlet is mainta
    6·1 answer
  • Two gases—neon and air—are expanded from P1 to P2 in a closed-system polytropic process with n = 1.2. _____ produces more work w
    7·1 answer
  • How to build a laser pointer?
    12·1 answer
  • One kilogram of air, initially at 5 bar, 350 K, and 3 kg of carbon dioxide (CO2), initially at 2 bar, 450 K, are confined to opp
    14·1 answer
  • A square isothermal chip is of width w 5 mm on a side and is mounted in a substrate such that its side and back surfaces are wel
    11·1 answer
  • A minor road intersects a major 4-lane divided road with a design speed of 50 mph and a median width of 12 ft. The intersection
    13·1 answer
  • Which of these is an example of a service job?
    7·1 answer
  • Diffrerentiate y=cos^{4} (3x+1)
    5·1 answer
  • The electron concentration in silicon at T = 300 K is given by
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!