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What is the difference between a refrigeration cycle and a heat pump cycle?

sukhopar [10]

Answer:

In refrigeration cycle heat transfer from inside refrigeration

In heat pump cycle heat transfer from environment

Explanation:

heat cycle is mechanical process use for cool the temperature but

In refrigeration heat transfer from inside of refrigeration that decrease temperature of refrigerator and in heat pump it decrease temperature negligible as compare to refrigerator

5 0

3 years ago

End A of the uniform 5-kg bar is pinned freely to the collar, which has an acceleration = 4 m/s2 along the fixed horizontal shaf

sergiy2304 [10]

Answer:

Fa = 57.32 N

Explanation:

given data

mass = 5 kg

acceleration = 4 m/s²

angular velocity ω = 2 rad/s

solution

first we take here moment about point A that is

∑Ma = Iα + ∑Mad ...............1

put here value and we get

so here I = ( $\frac{1}{12}$ ) × m × L² ................2

I = ( $\frac{1}{12}$ ) × 5 × 0.8²

I = 0.267 kg-m²

and

a is = r × α

a = 0.4 α

so now put here value in equation is 1

0 = 0.267 α + m r α (0.4) - m A (0.4)

0 = 0.267 α + 5 (0.4α) (0.4 ) - 5 (4) 0.4

so angular acceleration α = 7.5 rad/s²

so here force acting on x axis will be

∑ F(x) = m a(x) ..............3

a(x) = m a - m rα

put here value

a(x) = 5 × 4 - 5 × 0.4 × 7.5

a(x) = 5 N

and

force acting on y axis will be

∑ F(y) = m a(y) .............. 4

a(y) - mg = mrω²

a(y) - 5 × 9.81 = 5 × 0.4 × 2²

a(y) = 57.1 N

so

total force at A will be

Fa = $\sqrt{a(x)^2+a(y)^2}$ ...............5

Fa = $\sqrt{(57.1)^2+(5)^2}$

Fa = 57.32 N

3 0

3 years ago

Explain how collecting mean values (X-bar) from samples drawn from a dataset, regardless of the statistical distribution of the

Pani-rosa [81]

Answer:

answer is given below

Explanation:

Central Limit Theorem: The Central Limit Theorem (CLT) is a statistical theory that gives a sufficiently large sample size with limited variations from the population, the average of all samples from the same population is approximately the same. . In addition, all models follow a nearly normal distribution model.

The given phenomenon is described in the central limit theory. In other words, if we repeatedly take independent random samples of size n from any population, when n is large, the sample distribution is the normal distribution pattern.

mean of the sample means

$\mu _\bar x = \mu$ .............1

and here standard deviation of the sample means is

$\mu _\bar x = \frac{\sigma }{\sqrt{n}}$ .........2

and This theory is found elsewhere in the field of statistics. Although the central limit theory may seem abstract and devoid of any application, this theory is actually important for statistical practice.

8 0

3 years ago

A train which is traveling at 70 mi/hr applies its brakes as it reaches point A and slows down with a constant deceleration. Its

Ugo [173]

Answer:

a) 0 mi/s^2

b) 52 mi/s

Explanation:

Assuming the crossing is 1/2 mile past point A and that point B is near point A (it isn't clear in the problem)

The train was running at 70 mi/h at point A and with constant deceleration reachesn the crossing 1/2 mile away with a speed of 52 mi/h

The equation for position under constant acceleration is:

X(t) = X0 + V0 * t + 1/2 * a * t^2

I set my reference system so that the train passes point A at t=0 and point A is X = 0, so X0 = 0.

Also the equation for speed under constant acceleration is:

V(t) = V0 + a * t

Replacing

52 = 70 + a * t

Rearranging

a * t = 52 - 70

a = -18/t

I can then calculate the time it will take it to reach the crossing

1/2 * a * t^2 + V0 * t - X(t) = 0

Replacing

1/2 (-18/t) * t^ + 70 * t - 1/2 = 0

-9 * t + 70 * t = 1/2

61 * t = 1/2

t = (1/2)/61 = 0.0082 h = 29.5 s

And the acceleration is:

a = -18/0.0082 = -2195 mi/(h^2)

To beath the train the car must reach the crossing in 29.5 - 4.3 = 25.2 s

X(t) = X0 + V0 * t + 1/2 * a * t^2

52 mi/h = 0.0144 mi/s

1/2 = 0 + 0.0144 * 25.2 + 1/2 * a * 25.2^2

1/2 = 0.363 + 317.5 * a

317.5 * a = 0.5 - 0.363

a = 0.137/317.5 = 0.00043 mi/s^2 (its almost zero)

The car should remain at about constant speed.

It will be running at the same speed.

4 0

3 years ago

Write a program that uses a function called Output_Array_Info. Output_Array_Info Properties: Input Parameters: 1. A pointer to a

Artyom0805 [142]

Answer:

C++ code explained below

Explanation:

Please note the below program has been tested on ubuntu 16.04 system and compiled using g++ compiler. This code will also work on other IDE's

-----------------------------------------------------------------------------------------------------------------------------------

Program:

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//header files

#include<iostream>

//namespace

using namespace std;

//function defintion

void Output_Array_Info(int *array_ptr, int size)

{

//display all array elements

cout<<"Array elements are: "<<endl;

for(int i =0; i<size; i++)

{

cout<<*(array_ptr+i)<<endl;

}

//display address of each element

cout<<endl<<"memory address of each array elemnt is: "<<endl;

for(int i =0; i<size; i++)

{

cout<<array_ptr+i<<endl;

}

//start of main function

int main()

{

//pointer variables

int *pointer;

//an array

int numbers[] = { 5, 7, 9, 10, 12};

//pointer pointing to array

pointer = numbers;

//calculate the size of the array

int size = sizeof(numbers)/sizeof(int);

//call to function

Output_Array_Info(numbers, size);

return 0;

}

//end of the main program

8 0

3 years ago