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3 years ago

48/64 reduced to its lowest term

Engineering

2 answers:

QveST [7]3 years ago

5 0

Answer:

3/4

Explanation:

48/64 = 3/4

ArbitrLikvidat [17]3 years ago

4 0

The answer is 3/4

I just did it For homework Ok.
Bc V.

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5. Create a function named second_a that uses a list comprehension. The function will take a single integer parameter n. Find ev

Mekhanik [1.2K]

Answer:

//Program was implemented using C++ Programming Language

// Comments are used for explanatory purpose

#include<iostream>

using namespace std;

unsigned int second_a(unsigned int n)

{

int r,sum=0,temp;

int first;

for(int i= 1; I<=n; i++)

{

first = n;

//Check if first digit is 3

// Remove last digit from number till only one digit is left

while(first >= 10)

{

first = first / 10;

}

if(first == 3) // if first digit is 3

{

//Check if n is palindrome

temp=n; // save the value of n in a temporary Variable

while(n>0)

{

r=n%10; //getting remainder

sum=(sum*10)+r;

n=n/10;

}

if(temp==sum)

cout<<n<<" is a palindrome";

else

cout<<n<<" is not a palindrome";

}

Explanation:

The above code segments is a functional program that checks if a number that starts with digit 3 is Palindromic or not.

The program was coded using C++ programming language.

The main method of the program is omitted.

Comments were used for explanatory purpose.

8 0

3 years ago

Which type of modeling can create virtual designs that can save clients thousands of dollars?

swat32

Answer:

VR Prototyping

VR Prototyping Can Save you Thousands of Dollars.

Explanation:

there you go lad

8 0

3 years ago

The state of plane strain on an element is:

balu736 [363]

Answer:

a. ε₁=-0.000317

ε₂=0.000017

θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

$\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2} \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2} \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6} \pm 1.67 \times 10^{-4}$

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

$tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5$

θ= -13.28° and 76.72°

To determine the direction of ε₁ and ε₂

$\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2} + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2} + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56) + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\$

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain

$\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}$

=3.335 *10^-4

$\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )$

ε(avg) =150 *10^-6

orientation of γmax

$tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}$

θ = 31.71 or -58.29

To determine the direction of γmax

$\gamma _{x'y' }= - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }= - \frac{-300*10^{-6} - \ 0}{2} sin(63.42) + \frac{150*10^{-6}}{2}cos(63.42)$

= 1.67 *10^-4

4 0

4 years ago

An FPC 4 m2 in area is tested during the night to measure the overall heat loss coefficient. Water at 60 C circulates through th

sp2606 [1]

Answer:

<em> - 14.943 W/m^2K ( negative sign indicates cooling ) </em>

Explanation:

Given data:

Area of FPC = 4 m^2

temp of water = 60°C

flow rate = 0.06 l/s

ambient temperature = 8°C

exit temperature = 49°C

<u>Calculate the overall heat loss coefficient </u>

Note : heat lost by water = heat loss through convection

m*Cp*dT = h*A * ( T - To )

∴ dT / T - To = h*A / m*Cp ( integrate the relation )

In ( $\frac{49-8}{60-8}$ ) = h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )

In ( 41 / 52 ) = 0.0159*h

hence h = - 0.2376 / 0.0159

= - 14.943 W/m^2K ( heat loss coefficient )

7 0

3 years ago

Technician A says the final drive assembly always has a gear ratio of 1:1. Technician B says the final drive assembly provides f

Olenka [21]

Answer:

Technician B only is correct

Explanation:

The last stage of gears found between the vehicle transmission system and the wheels is the final drive ratio. The function of the final drive gear assembly is to enable a gear reduction control stage to reduce the rotation per minute and increase the wheel torque, such that the vehicle performance can be adjusted and the final gear ratio can be between 3:1 and 4.5:1 not 1:1

Therefore, technician B only is correct

5 0

3 years ago