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3 years ago

When the psychologist simply records the relationship between two variables...

1 answer:

8 0

When a psychologist simply records the relationship between two variables without manipulating them, it is called a correlational study.

The observed relationship does not by itself reveal which variable causes the other. This is the directionally problem. Also, the relationship may be due to a third variable controlling both of the observed variables.

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The wheel and the attached reel have a combined weight of 50lb and a radius of gyration about their center of 6 A k in = . If pu

marishachu [46]

The complete question is;

The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of ka = 6 in. If pulley B that is attached to the motor is subjected to a torque of M = 50 lb.ft, determine the velocity of the 200lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.

The image of this system is attached.

Answer:

Velocity = 11.8 ft/s

Explanation:

Since the wheel at A rotates about a fixed axis, then;

v_c = ω•r_c

r_c is 4.5 in. Let's convert it to ft.

So, r_c = 4.5/12 ft = 0.375 ft

Thus;

v_c = 0.375ω

Now the mass moment of inertia about of wheel A about it's mass centre is given as;

I_a = m•(k_a)²

The mass in in lb, so let's convert to slug. So, m = 50/32.2 slug = 1.5528 slug

Also, let's convert ka from inches to ft.

So, ka = 6/12 = 0.5

So,I_a = 1.5528 × 0.5²

I_a = 0.388 slug.ft²

The kinetic energy of the system would be;

T = Ta + Tc

Where; Ta = ½•I_a•ω²

And Tc = ½•m_c•(v_c)²

So, T = ½•I_a•ω² + ½•m_c•(v_c)²

Now, m_c is given as 200 lb.

Converting to slug, we have;

m_c = (200/32.2) slugs

Plugging in the relevant values, we have;

T = (½•0.388•ω²) + (½•(200/32.2)•(0.375ω)²)

This now gives;

T = 0.6307 ω²

The system is initially at rest at T1 = 0.

Resolving forces at A, we have; Ax, Ay and Wa. These 3 forces do no work.

Whereas at B, M does positive work and at C, W_c does negative work.

When pulley B rotates, it has an angle of; θ_b = 5 revs × 2π rad/revs = 10π

While the wheel rotates through an angle of;θ_a = (rb/ra) • θ_b

Where, rb = 3 in = 3/12 ft = 0.25 ft

ra = 7.5 in = 7.5/12 ft = 0.625 ft

So, θ_a = (0.25/0.625) × 10π

θ_a = 4π

Thus, we can say that the crate will have am upward displacement through a distance;

s_c = r_c × θ_a = 0.375 × 4π

s_c = 1.5π ft

So, the work done by M is;

U_m = M × θ_b

U_m = 50lb × 10π

U_m = 500π

Also,the work done by W_c is;

U_Wc = -W_c × s_c = -200lb × 1.5π

U_Wc = -300π

From principle of work and energy;

T1 + (U_m + U_Wc) = T

Since T1 is zero as stated earlier,

Thus ;

0 + 500π - 300π = 0.6307 ω²

0.6307ω² = 200π

ω² = 200π/0.6307

ω² = 996.224

ω = √996.224

ω = 31.56 rad/s

We earlier derived that;v_c = 0.375ω

Thus; v_c = 0.375 × 31.56

v_c = 11.8 ft/s

3 0

3 years ago

An AX ceramic compound has the rock salt crystal structure. If the radii of the A and X ions are 0.137 and 0.241 nm, respectivel

Tju [1.3M]

Answer:

c) 1.75 g/cm³

Explanation:

Given that

Radii of the A ion, r(c) = 0.137 nm

Radii of the X ion, r(a) = 0.241 nm

Atomic weight of the A ion, A(c) = 22.7 g/mol

Atomic weight of the X ion, A(a) = 91.4 g/mol

Avogadro's number, N = 6.02*10^23 per mol

Solution is attached below

3 0

3 years ago

Wireless technology has made communicating

emmainna [20.7K]

Answer:

Taking as a basis of calculation 100 mol of gas leaving the conversion reactor, draw andcompletely label a flowchart of this process. Then calculate the moles of fresh methanol feed,formaldehyde product solution, recycled methanol, and absorber off-gas, the kg of steamgenerated in the waste-heat boiler, and the kg of cooling water fed to the heat exchangerbetween the waste-heat boiler and the absorber. Finally, calculate the heat (kJ) that must beremoved in the distillation column overhead condenser, assuming that methanol enters as asaturated vapor at 1 atm and leaves as a saturated liquid at the same pressure.

SEE ANSWER

Explanation:

6 0

3 years ago

A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

Bad White [126]

Answer:

the net work per cycle $\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

diameter of the four-cylinder bore = 3.7 in

length of the stroke = 3.4 in

The clearance volume = 16% = 0.16

The cylindrical volume $V_2 = 0.16 V_1$

the crankshaft N rotates at a speed of 2400 RPM.

At the beginning of the compression , temperature $T_1$ = 60 F = 519.67 R

and;

Otto cycle with a pressure = 14.5 lbf/in² = (14.5 × 144 ) lb/ft²

= 2088 lb/ft²

The maximum temperature in the cycle is 5200 R

From the given information; the change in volume is:

$V_1-V_2 = \dfrac{\pi}{4}D^2L$

$V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)$

$V_1-0.16V_1= 36.55714291$

$0.84 V_1 =36.55714291$

$V_1 =\dfrac{36.55714291}{0.84 }$

$V_1 =43.52040823 \ in^3 \\ \\ V_1 = 43.52 \ in^3$

$V_1 = 0.02518 \ ft^3$

the mass in air ( lb) can be determined by using the formula:

$m = \dfrac{P_1V_1}{RT}$

where;

R = 53.3533 ft.lbf/lb.R°

$m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}$

m = 0.0018962 lb

From the tables of ideal gas properties at Temperature 519.67 R

$v_{r1} =158.58$

$u_1 = 88.62 Btu/lb$

At state of volume 2; the relative volume can be determined as:

$v_{r2} = v_{r1} \times \dfrac{V_2}{V_1}$

$v_{r2} = 158.58 \times 0.16$

$v_{r2} = 25.3728$

The specific energy $u_2$ at $v_{r2} = 25.3728$ is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

$v_{r3} = 0.1828$

$u_3 = 1098 \ Btu/lb$

To determine the relative volume at state 4; we have:

$v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}$

$v_{r4} =0.1828 \times \dfrac{1}{0.16}$

$v_{r4} =1.1425$

The specific energy $u_4$ at $v_{r4} =1.1425$ is 591.84 Btu/lb

Now; the net work per cycle can now be calculated as by using the following formula:

$W_{net} = Heat \ supplied - Heat \ rejected$

$W_{net} = m(u_3-u_2)-m(u_4 - u_1)$

$W_{net} = m(u_3-u_2- u_4 + u_1)$

$W_{net} = m(1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (410.08)$

$\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, in horsepower. can be calculated as follows;

In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:

$W = 4 \times N' \times W_{net$

where ;

$N' = \dfrac{2400}{2}$

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec × 0.777593696

W = 62.20749568 Btu/s

W = 88.0144746 hp

8 0

3 years ago

Given a matrix, clockwise-rotate elements in it. Please add code to problem3.cpp and the makefile. Use the code in p3 to test yo

rusak2 [61]

Answer:

/* C Program to rotate matrix by 90 degrees */

#include<stdio.h>

int main()

{

int matrix[100][100];

int m,n,i,j;

printf("Enter row and columns of matrix: ");

scanf("%d%d",&m,&n);

/* Enter m*n array elements */

printf("Enter matrix elements: \n");

for(i=0;i<m;i++)

{

for(j=0;j<n;j++)

{

scanf("%d",&matrix[i][j]);

}

/* matrix after the 90 degrees rotation */

printf("Matrix after 90 degrees roration \n");

for(i=0;i<n;i++)

{

for(j=m-1;j>=0;j--)

{

printf("%d ",matrix[j][i]);

}

printf("\n");

}

return 0;

}

5 0

3 years ago