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3 years ago

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Min is conducting an experiment where he compares the properties of water and lemonade. The first stage of the experiment is foc

using only on the physical properties of the water and lemonade. Which comparison point would Min not use in the first stage?

1 answer:

madam [21]3 years ago

5 0

Answer:

only if i knew

Explanation:

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Write the implementation (.cpp file) of the Player class from the previous exercise. Again, the class contains:

hoa [83]

Answer:

//Define the header file

#ifndef PLAYER_H

#define PLAYER_H

//header file.

#include <string>

//Use the standard namespace.

using namespace std;

//Define the class Player.

class Player

{

//Declare the required data members.

string name;

int score;

public:

//Declare the required

//member functions.

void setName(string par_name);

void setScore(int par_score);

string getName();

int getScore();

}

//End the definition

//of the header file.

#endif

Player.cpp:

//Include the "Player.h" header file,

#include "Player.h"

//Define the setName() function.

void Player::setName(string par_name)

{

name = par_name;

}

//Define the setScore() function.

void Player::setScore(int par_score)

{

score = par_score;

}

//Define the getName() function.

string Player::getName()

{

return name;

}

//Define the getScore() function.

int Player::getScore()

{

return score;

}

7 0

3 years ago

In a planetary geartrain with a form factor of 8, the sun gear rotates clockwise at 5 rad⁄s and the ring gear rotates clockwise

lina2011 [118]

Answer:

D. N= 11. 22 rad/s (CW)

Explanation:

Given that

Form factor R = 8

Speed of sun gear = 5 rad/s (CW)

Speed of ring gear = 12 rad/s (CW)

Lets take speed of carrier gear is N

From Algebraic method ,the relationship between speed and form factor given as follows

$\dfrac{N_{sun}-N}{N_{ring}-N}=-R$

here negative sign means that ring and sun gear rotates in opposite direction

Lets take CW as positive and ACW as negative.

Now by putting the values

$\dfrac{N_{sun}-N}{N_{ring}-N}=-R$

$\dfrac{5-N}{12-N}=-8$

N= 11. 22 rad/s (CW)

So the speed of carrier gear is 11.22 rad/s clockwise.

8 0

3 years ago

Calculate the unit cell edge length for an 80 wt% Ag−20 wt% Pd alloy. All of the palladium is in solid solution, the crystal str

alisha [4.7K]

Answer:

λ^3 = 4.37

Explanation:

first let us to calculate the average density of the alloy

for simplicity of calculation assume a 100g alloy

80g --> Ag

20g --> Pd

ρ_avg= 100/(20/ρ_Pd+80/ρ_avg)

= 100*10^-3/(20/11.9*10^6+80/10.44*10^6)

= 10744.62 kg/m^3

now Ag forms FCC and Pd is the impurity in one unit cell there is 4 atoms of Ag since Pd is the impurity we can not how many atom of Pd in one unit cell let us calculate

total no of unit cell in 100g of allow = 80 g/4*107.87*1.66*10^-27

= 1.12*10^23 unit cells

mass of Pd in 1 unit cell = 20/1.12*10^23

Now,

ρ_avg= mass of unit cell/volume of unit cell

ρ_avg= (4*107.87*1.66*10^-27+20/1.12*10^23)/λ^3

λ^3 = 4.37

6 0

3 years ago

A cylindrical buoy is 2m in diameter and 2.5m long and weight 22kN . The specific weight of sea water is 10.25kN/m^3 . (I) Show

aleksandrvk [35]

Answer:

$GM$

So the bouy does not float with its axis vertical

Explanation:

From the question we are told that:

Diameter $d=2m$

Length $l=2.5m$

Weight $W=22kN$

Specific weight of sea water $\mu= 10.25kN/m^3$

Generally the equation for weight of cylinder is mathematically given by

Weight of cylinder = buoyancy Force

$W=(pwg)Vd$

Where

$V_d=\pi/4(d)^2y$

Therefore

$22*10^3=10.25*10^3 *\pi/4(2)^2y\\\\\22*10^3=32201.3247y\\\\\y=1.5m$

Therefore

Center of Bouyance B

$B=\frac{y}{2}=0.26m\\\\B=0.75$

Center of Gravity

$G=\frac{I.B}{2}=2.6m$

Generally the equation for\BM is mathematically given by

$BM=\frac{I}{vd}\\\\BM=\frac{3.142/64*2^4}{3.142/4*2^2*0.5215}\\\\BM=0.479m\\\\$

Therefore

$BG=2.6-0.476\\\\BG=0.64m$

Therefore

$GM=BM-BG\\\\GM=0.479m-0.64m\\\\GM=-0.161m\\\\$

Therefore

$GM$

So the bouy does not float with its axis vertical

4 0

3 years ago

In 1945, the United States tested the world’s first atomic bomb in what was called the Trinity test. Following the test, images

Zarrin [17]

Answer:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

$A= \frac{r^5 \rho}{t^2}$

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

Explanation:

Notation

In order to do the dimensional analysis we need to take in count that we need to conditions:

a) The energy A is released in a small place

b) The shock follows a spherical pattern

We can assume that the size of the explosion r is a function of the time t, and depends of A (energy), the time (t) and the density of the air is constant $\rho_{air}$ .

And now we can solve the dimensional problem. We assume that L is for the distance T for the time and M for the mass.

[r]=L with r representing the radius

[A]= $\frac{ML^2}{T^2}$ A represent the energy and is defined as the mass times the velocity square, and the velocity is defined as $\frac{L}{T}$

[t]=T represent the time

$[\rho]=\frac{M}{L^3}$ represent the density.

Solution to the problem

And if we analyze the function for r we got this:

$[r]=L=[A]^x [\rho]^y [t]^z$

And if we replpace the formulas for each on we got:

$[r]=L =(\frac{ML^2}{T^2})^x (\frac{M}{L^3})^y (T)^z$

And using algebra properties we can express this like that:

$[r]=L=M^{x+y} L^{2x-3y} T^{-2x+z}$

And on this case we can use the exponents to solve the values of x, y and z. We have the following system.

$x+y =0 , 2x-3y=1, -2x+z=0$

We can solve for x like this x=-y and replacing into quation 2 we got:

$2(-y)-3y = 1$

$-5y = 1$

$y= -\frac{1}{5}$

And then we can solve for x and we got:

$x = -y = -(-\frac{1}{5})=\frac{1}{5}$

And if we solve for z we got:

$z=2x =2 \frac{1}{5}=\frac{2}{5}$

And now we can express the radius in terms of the dimensional analysis like this:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

And K represent a constant in order to make the porportional relation and equality.

The problem says that we can assume the constant K=1.

And if we solve for the energy we got:

$A^{1/5}=\frac{r}{t^{2/5} \rho^{-1/5}}$

$A= \frac{r^5 \rho}{t^2}$

And now we can replace the values given. On this case t =0.025 s, the radius r =140 m, and the density is a constant assumed $\rho =1.2 kg/m^2$ , and replacing we got:

$A=\frac{140^5 1.2 kg/m^3}{(0.025 s)^2}=1.033x10^{14} \frac{kg m^2}{s^2}$

And we can convert this into ergs we got:

$A= 1.033x10^{14} \frac{kgm^2}{s^2} * \frac{1 x10^7 egrs}{1 \frac{kgm^2}{s^2}}=1.033x10^{21} ergs$

And then we know that 1 g of TNT have $4x10^4 erg$

And we got:

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

3 0

3 years ago