other examples of joption​

Answers

Answer 1

Answer:

Profit from stock price gains with limited risk and lower cost than buying the stock outright.

Profit from stock price drops with limited risk and lower cost than shorting the stock.

Profit from sideways markets by selling options and generating income.

Get paid to buy stock.

Explanation:

I don’t know if you meant option but I hope this helps.


Related Questions

A heat pump operates on a vapor-compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the compressor as saturated vapor at X kPa, and exits at 800 kPa and 60oC. The refrigerant leaves the condenser as saturated liquid at 800 kPa. The number of letters in your first name multiplied by 10 plus 50

Answers

Answer:

Hello your question has some missing information below are the missing information

The refrigerant enters the compressor as saturated vapor at 140kPa Determine The coefficient of performance of this heat pump

answer : 2.49

Explanation:

For  vapor-compression refrigeration cycle

P1 = P4  ; P1 = 140 kPa

P2( pressure at inlet ) = P3 ( pressure at outlet ) ; P2 = 800 kPa

From pressure table of R 134a refrigerant

h1 ( enthalpy of saturated vapor at 140kPa ) = 239.16 kJ/kg

h2 ( enthalpy of saturated liquid at P2 = 800 kPa and t = 60°C )

= 296.8kJ/kg

h3 ( enthalpy of saturated liquid at P3 = 800 kPa ) = 95.47 kJ/kg

also h4 = 95.47 kJ/kg

To determine the coefficient of performance  

Cop = ( h1 - h4 ) / ( h2 - h1 )

∴ Cop = 2.49

A chemical process stream enters a shell-and-tube exchanger at a temperature of 200.0°Fand does two passes on the shell side, exiting the exchange at 170.0°F. The Heat exchanger Spring 2021 has 200 stainless steel tubes that are 2-in.ODand 10.0ft long. Indicate whether the temperature of the process stream will increase, decrease, or remain the sameunder the following scenarios. You must justify your answerto receive full credit.a)The flow rate of the cooling fluid is increased.b)There are 200 tubes that are 1.0-in. OD and 20.0ft long.c)The number of shell passes is doubled.d)The tube material is changed to copper.

Answers

Answer:

a) Decrease

b) Decrease

c) Decrease

d) Decrease

Explanation:

Ti= 200°F ,

Te =  170°F

Area of heat exchanger = [tex]\pi *(2 )* 10[/tex]  = 20π

A) when The flow rate of the cooling fluid is increased

Temperature of process stream will decrease  and this is because the tube side heat transfer coefficient will increase and this will increase the rate of heat transfer thereby decreasing the temperature of the process stream.

B) when There are 200 tubes that are 1.0-in. OD and 20.0ft long

The temperature of the process stream will decrease and this is because the heat transfer coefficient will increase likewise the heat transfer rate

C) When The number of shell passes is doubled

This will cause an increase in the overall length of the shell, an increase in velocity of constant volumetric flowrate, hence the Temperature of the process steam will decrease as well

D) When The tube material is changed to copper.

Due to the high thermal conductivity of copper when compared to steel , switching to copper will cause a decrease in the temperature of the process steam

a) The initial moisture content of a food product is 77% (wet basis), and the critical moisture content is 30% (wet basis). If the constant drying rate in a fluidized bed dryer is 0.1 kg water removed/m2-s, determine the time required for the product to begin the falling-rate drying period. The product has a cube shape with 5-cm sides; the initial product density is 950 kg/m3.

Answers

Answer:

≈ 53 seconds

Explanation:

calculate Time required for the product to begin the falling-rate drying period

Initial moisture content =  0.77 kg water /kg of product

                                               = 3.35 kg water /kg solids

Critical moisture content = 0.3 kg water / kg product

                                         = 0.43 kg water / kg solids

∴ amount of water to be removed = 3.35 - 0.43 = 2.95kg water /kg solids

next: calculate surface are a of product during drying

= (0.05 * 0.05 ) * 6

= 0.015 m^3

Drying rate = 0.1 kg water m^2.s^-1 * 0.015 m^3 = 1.5 * 10^-3 kg water s^-1

applying product density

initial product mass = 0.11875 * 0.23 = 0.0273kg solid

hence total amount of water to removed = 2.92 * 0.0273 = 0.07972 kg

therefore : Time required for the product to begin the falling-rate drying period

= 0.07972 / 1.5 * 10^-3

= 53 seconds

An interest rate of norminal 12% per year , compounded weekly is

Answers

Answer: It is  a nominal rate per year

what technology has been used for building super structures​

Answers

Answer: Advanced technologixal machines

Explanation: such as big cranes, multiple workers helping creat said structure, and big bull dozers

Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 20oC and exits at 3.0 bar. The refrigerant undergoes a throttling process. Determine the temperature, in oC, and the quality of the refrigerant at the exit of the expansion valve. Step 1 Determine the temperature of the refrigerant at the exit, in oC.

Answers

Answer:

[tex]T_{2}[/tex] = -9.24 °C

x = 0.1057

Explanation:

The tables used in this answer and explanation come from Fundamentals of Engineering Thermodynamics 9th Edition.

Using Table A-14: Properties of Saturated Ammonia (Liquid-Vapor): Pressure Table and the given [tex]P_{2}[/tex], [tex]T_{2}[/tex] can be determined by finding the temperature that corresponds with [tex]P_{2}[/tex] on the table. In this case, [tex]T_{2}[/tex] = -9.24 °C.

The quality of the refrigerant can be determined by using data from the same table and [tex]h_{2} =274.26[/tex] kJ/kg.

Necessary data (P=3bar):

[tex]h_{f}=137.42[/tex] kJ/kg

[tex]h_{g}=1431.47[/tex] kJ/kg

The formula to calculate quality is [tex]h_{2} =h_{f}+x(h_{g}-h_{f})[/tex].

Rearranging for x:

[tex]x=\frac{h_{2}-h_{f} }{h_{g}-h_{f} }= \frac{274.26-137.42}{1431.47-137.42}=0.1057[/tex]

Q5) Write C++ program to find the summation of sines of the even values that can be divided by 7 between -170 and -137.

Answers

Answer:

In C++:

#include <iostream>

#include <cmath>

using namespace std;

int main(){

 double total = 0;

 for(int i = -170; i<=-130;i++){

     if(i%2 == 0 && i%7==0){

         double angle = i*3.14159/180;

         total+=sin(angle);      }  }

 cout<<"Total: "<<total;

 return 0; }

Explanation:

This initializes the total to 0

 double total = 0;

This iterates from -170 to - 130

 for(int i = -170; i<=-130;i++){

This checks if the current number is even and is divided by 7

     if(i%2 == 0 && i%7==0){

This converts the number from degrees to radians

         double angle = i*3.14159/180;

This adds the sine of the number

         total+=sin(angle);      }  }

This prints the calculated total

 cout<<"Total: "<<total;

brainly and points if you want

Answers

Answer:

thank you

Explanation:

have a nice day

Answer:

thankd

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