What is the smallest variable type I can use to represent the number 27?

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

Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 8-mm-diameter, 40-m-long horiz

Naddik [55]

Answer:

Q = 5.06 x 10⁻⁸ m³/s

Explanation:

Given:

v=0.00062 m² /s and ρ= 850 kg/m³

diameter = 8 mm

length of horizontal pipe = 40 m

Dynamic viscosity =

μ = ρv

=850 x 0.00062

= 0.527 kg/m·s

The pressure at the bottom of the tank is:

P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²

The laminar flow rate through a horizontal pipe is:

$Q = \dfrac{\Delta P \pi D^4}{128 \mu L}$

$Q= \dfrac{33.32 \times 1000 \pi\times 0.008^4}{128 \times 0.527 \times 40}$

Q = 5.06 x 10⁻⁸ m³/s

4 0

2 years ago

Fix the code so the program will run correctly for MAXCHEESE values of 0 to 20 (inclusive). Note that the value of MAXCHEESE is

GarryVolchara [31]

Answer:

Code fixed below using Java

Explanation:

<u>Error.java </u>

import java.util.Random;

public class Error {

public static void main(String[] args) {

final int MAXCHEESE = 10;

String[] names = new String[MAXCHEESE];

double[] prices = new double[MAXCHEESE];

double[] amounts = new double[MAXCHEESE];

// Three Special Cheeses

names[0] = "Humboldt Fog";

prices[0] = 25.00;

names[1] = "Red Hawk";

prices[1] = 40.50;

names[2] = "Teleme";

prices[2] = 17.25;

System.out.println("We sell " + MAXCHEESE + " kind of Cheese:");

System.out.println(names[0] + ": $" + prices[0] + " per pound");

System.out.println(names[1] + ": $" + prices[1] + " per pound");

System.out.println(names[2] + ": $" + prices[2] + " per pound");

Random ranGen = new Random(100);

// error at initialising i

// i should be from 0 to MAXCHEESE value

for (int i = 0; i < MAXCHEESE; i++) {

names[i] = "Cheese Type " + (char) ('A' + i);

prices[i] = ranGen.nextInt(1000) / 100.0;

amounts[i] = 0;

System.out.println(names[i] + ": $" + prices[i] + " per pound");

}

7 0

3 years ago

In normal operation, a paper mill generates excess steam at 20 bar and 400◦C. It is planned to use this steam as the feed to a t

Keith_Richards [23]

Answer:

The maximum power that can be generated is 127.788 kW

Explanation:

Using the steam table

Enthalpy at 20 bar = 2799 kJ/kg

Enthalpy at 2 bar = 2707 kJ/kg

Change in enthalpy = 2799 - 2707 = 92 kJ/kg

Mass flow rate of steam = 5000 kg/hr = 5000 kJ/hr × 1 hr/3600 s = 1.389 kg/s

Maximum power generated = change in enthalpy × mass flow rate = 92 kJ/kg × 1.389 kg/s = 127.788 kJ/s = 127.788 kW

6 0

3 years ago

What’s the difference between engineering stress and strain and true stress and strain

Nana76 [90]

True strain and engineering strain? True stress is defined as the load divided by the cross-sectional area of the specimen at that instant and is a true indication of the internal pressures. ... Engineering stress is defined as the load divided by the initial cross-sectional area of the specimenAnswer:

Explanation:

4 0

3 years ago

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