If you are driving a 30-foot

Two Carnot engines operate in series such that the heat rejected from one is the heat input to the other. The heat transfer from

kykrilka [37]

Answer:

Given:

high temperature reservoir $T_{H} =1000k$

low temperature reservoir $T_{L} =400k$

thermal efficiency $n_{1}= n_{2}$

The engines are said to operate on Carnot cycle which is totally reversible.

To find the intermediate temperature between the two engines, The thermal efficiency of the first heat engine can be defined as

$n_{1} =1-\frac{T}{T_{H} }$

The thermal efficiency of second heat engine can be written as

$n_{2} =1-\frac{T_{L} }{T}$

The temperature of intermediate reservoir can be defined as

$1-\frac{T}{T_{H} } =1-\frac{T_{L} }{T} \\T^2=T_{L} T_{H} \\T=\sqrt{T_{L} T_{H} }\\T=\sqrt{400*1000} =632k$

8 0

3 years ago

Create an abstract class DiscountPolicy. It should have a single abstract method computeDiscount that will return the discount f

eimsori [14]

Answer:

Java Code was used to define classes in the abstract discount policy,The bulk discount, The buy items get one free and the combined discount

Explanation:

Solution

Code:

Main.java

public class Main {

public static void main(String[] args) {

BulkDiscount bd=new BulkDiscount(10,5);

BuyNItemsGetOneFree bnd=new BuyNItemsGetOneFree(5);

CombinedDiscount cd=new CombinedDiscount(bd,bnd);

System.out.println("Bulk Discount :"+bd.computeDiscount(20, 20));

System.out.println("Nth item discount :"+bnd.computeDiscount(20, 20));

System.out.println("Combined discount :"+cd.computeDiscount(20, 20));

}

discountPolicy.java

public abstract class DiscountPolicy

{

public abstract double computeDiscount(int count, double itemCost);

}

BulkDiscount.java

public class BulkDiscount extends DiscountPolicy

{

private double percent;

private double minimum;

public BulkDiscount(int minimum, double percent)

{

this.minimum = minimum;

this.percent = percent;

}

at Override

public double computeDiscount(int count, double itemCost)

{

if (count >= minimum)

{

return (percent/100)*(count*itemCost); //discount is total price * percentage discount

}

return 0;

}

BuyNItemsGetOneFree.java

public class BuyNItemsGetOneFree extends DiscountPolicy

{

private int itemNumberForFree;

public BuyNItemsGetOneFree(int n)

{

itemNumberForFree = n;

}

at Override

public double computeDiscount(int count, double itemCost)

{

if(count > itemNumberForFree)

return (count/itemNumberForFree)*itemCost;

else

return 0;

}

CombinedDiscount.java

public class CombinedDiscount extends DiscountPolicy

{

private DiscountPolicy first, second;

public CombinedDiscount(DiscountPolicy firstDiscount, DiscountPolicy secondDiscount)

{

first = firstDiscount;

second = secondDiscount;

}

at Override

public double computeDiscount(int count, double itemCost)

{

double firstDiscount=first.computeDiscount(count, itemCost);

double secondDiscount=second.computeDiscount(count, itemCost);

if(firstDiscount>secondDiscount){

return firstDiscount;

}else{

return secondDiscount;

}

6 0

3 years ago

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

What are the atomic binding force and energy? how do they relate to materials strength and thermal stability.

Elanso [62]

Answer:

As we know that every molecule is attached by a strong force .The force required to disassemble the atoms is know as atomic binding force or we can say that the force required to disassemble the electron from atoms is known as binding force.On the other hand the energy require to doing this is known as atomic binding energy.

If the binding force is high then it will become difficult to disassemble thermally as well as mechanically.So we can say that it have direct relationship with materials strength and thermal stability.

7 0

3 years ago

Which of the following is NOT an example of a direct cost of workplace injuries?

hammer [34]

Answer:

Lost productivity

4 0

3 years ago