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2 years ago

Without motorcycle riders are at risk of severe injury in a crash ?

Engineering

2 answers:

astra-53 [7]2 years ago

6 0

The answer is A
Proper protection

nekit [7.7K]2 years ago

5 0

Answer:

The correct option is;

A. proper protection

Explanation:

Motorcycle riders ride the motorcycle while at some level of speed while having the entire body exposed to be a major part of any collision.

Injuries sustained from motorcycle accidents are several times more severe than injuries sustained by occupants of a car that is fully protected by the metallic panel in the same and even more serious accident scenarios

Hence, motorcycle riders require adequate protection by putting on available motorcyclist safety gear

Therefore, to reduce the risk of severe injury n a crash, motorcycle riders require proper protection.

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A 1000 W iron utilizes a resistance wire which is 20 inches long and has a diameter of 0.08 inches. Determine the rate of heat g

SSSSS [86.1K]

Answer:

The rate of heat generation in the wire per unit volume is 5.79×10^7 Btu/hrft^3

Heat flux is 9.67×10^7 Btu/hrft^2

Explanation:

Rate of heat generation = 1000 W = 1000/0.29307 = 3412.15 Btu/hr

Area (A) = πD^2/4

Diameter (D) = 0.08 inches = 0.08 in × 3.2808 ft/39.37 in = 0.0067 ft

A = 3.142×0.0067^2/4 = 3.53×10^-5 ft^2

Volume (V) = A × Length

L = 20 inches = 20 in × 3.2808 ft/39.37 in = 1.67 ft

V = 3.53×10^-5 × 1.67 = 5.8951×10^-5 ft^3

Rate of heat generation in the wire per unit volume = 3412.15 Btu/hr ÷ 5.8951×10^-5 ft^3 = 5.79×10^7 Btu/hrft^3

Heat flux = 3412.15 Btu/hr ÷ 3.53×10^-5 ft^2 = 9.67×10^7 Btu/hrft^2

3 0

2 years ago

Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr

son4ous [18]

Answer:

The entropy change of the air is $0.240kJ/kgK$

Explanation:

$T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa$

$T_{2}$ is unknown

we can apply the following expression to find $T_{2}$

$-w_{out} =mc_{v} (T_{2} -T_{1} )$

$T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }$

now substitute

$T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K$

To find entropy change of the air we can apply the ideal gas relationship

Δ $s_{air}$ $=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }$

Δ $s_{air} =$ $1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )$

Δ $s_{air} =0.240kJ/kgK$

4 0

2 years ago

LAB 3.3 – Working with String Input and Type CastingStep 1: RemovefindErrors.cppfrom the project and add thepercentage.cppprogra

jolli1 [7]

Answer:

// Program is written in C++ Programming Language

// Comments are used for explanatory purpose

#include<iostream>

using namespace std;

int main ()

{

// Variable declaration

string name;

int numQuestions;

int numCorrect;

double percentage;

//Prompt to enter student's first and last name

cout<<"Enter student's first and last name";

cin>>name; // this line accepts input for variable name

cout<<"Number of question on test"; //Prompt to enter number of questions on test

cin>> numQuestions; //This line accepts Input for Variable numQuestions

cout<<"Number of answers student got correct: "; // Prompt to enter number of correct answers

cin>>numCorrect; //Enter number of correct answers

percentage = numCorrect * 100 / numQuestions; // calculate percentage

cout<<name<<" "<<percentage<<"%"; // print

return 0;

}

Explanation:

The code above calculates the percentage of a student's score in a certain test.

The code is extracted from the Question and completed after extraction.

It's written in C++ programming language

4 0

3 years ago

The ???? − i relationship for an electromagnetic system is given by ???? = 1.2i1/2 g where g is the air-gap length. For current

Artemon [7]

Answer:

a) The mechanical force is -226.2 N

b) Using the coenergy the mechanical force is -226.2 N

Explanation:

a) Energy of the system:

$\lambda =\frac{1.2*i^{1/2} }{g} \\i=(\frac{\lambda g}{1.2} )^{2}$

$\frac{\delta w_{f} }{\delta g} =\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }$

$f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }$

If i = 2A and g = 10 cm

$\lambda =\frac{1.2*i^{1/2} }{g} =\frac{1.2*2^{1/2} }{10x10^{-2} } =16.97$

$f_{m}=-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }=-\frac{16.97^{3}*2*0.1 }{3*1.2^{2} } =-226.2N$

b) Using the coenergy of the system:

$f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{1.2*2*i^{3/2} }{3*g^{2} }=-\frac{1.2*2*2^{3/2} }{3*0.1^{2} } =-226.2N$

8 0

2 years ago

What is this spray pattern defect most likely caused by:

DiKsa [7]

Answer:

fluid nozzle that is too large

6 0

1 year ago