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3 years ago

10

Make a sketch of a simple mechanically expanded brake and indicate the forces acting on the leading shoe when the brake is a

pplied.

1 answer:

Pavel [41]3 years ago

4 0

Answer: sorry I just need points

Explanation

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What is the sun's degree angle in the sky in summer and in winter?

Vladimir [108]

Answer:

At the time of the summer or winter solstices, the Sun is 23.44° degrees above or below the horizon, respectively, irrespective of time of day.

Explanation:

6 0

2 years ago

A heat pump with an ideal compressor operates between 0.2 MPa and 1 MPa. Refrigerant R134a flows through the system at a rate of

solmaris [256]

Answer:

The mass flow rate of refrigerant is 0.352 kg/s

Explanation:

Considering the cycle of an ideal heat pump, provided in the attachment, we first find enthalpy at state B and D. For that purpose, we use property tables of refrigerant R134a:

<u>At State A</u>:

From table, we see the enthalpy and entropy value of saturated vapor at 0.2 MPa. Therefore:

ha = 244.5 KJ/kg

Sa = 0.93788 KJ/kg.k

<u>At State B</u>:

Since, the process from state A to B is isentropic. Therefore,

Sb = Sa = 0.93788 KJ/Kg

From table, we see the enthalpy value of super heated vapor at 1 MPa and Sb. Therefore:

hb = 256.85 KJ/kg (By interpolation)

<u>At State C</u>:

From table, we see the enthalpy and entropy value of saturated liquid at 1 MPa. Therefore:

hc = 107.34 KJ/kg

Now, from the diagram it is very clear that:

Heat Loss = m(hb = hc)

m = (Heat Loss)/(hb - hc)

where,

m = mass flow rate = ?

Heat Loss = (180,000 Btu/hr)(1.05506 KJ/1 Btu)(1 hr/3600 sec)

Heat Loss = 52.753 KW

Therefore,

m = (52.753 KJ/s)/(256.85 KJ/kg - 107.34 KJ/kg)

<u>m = 0.352 kg/s</u>

5 0

3 years ago

C++ - Green Crud Fibonacci programThe following program is to be written with a loop. You are to write this program three times

Fynjy0 [20]

Answer:

Below is the required code:

Explanation:

Using for loop

#include <iostream>

using namespace std;

int main()

{

//Initial crud size

int init = 0;

int newCrud;

int next=0;

//Number of days to simulate

int no_days;

int day;

cout << "Enter initial amount of green crud: ";

cin >> newCrud;

cout << "Enter number of days to simulate: ";

cin >> no_days;

for (day = 10; day<=no_days; day++)

{

if (day % 10 == 0)

{

next = newCrud + init;

}

newCrud = init;

init = next;

}

if (no_days < 5)

cout << "\nCrud reproduce only after 5 days minimum.Hence the current amount is "

<< newCrud << " pounds.";

else

cout << "On day " << no_days << " you have " << init

<< " pounds of green crud." << endl;

cout << "\nWould you like to continue? (y or n): ";

cin >> ans;

return 0;

}

Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

Using while loop

Program:

#include <iostream>

using namespace std;

int main()

{

char ans='y';

while (ans == 'Y' || ans == 'y')

{

//Initial crud size

int init = 0;

int newCrud;

int next=0;

//Number of days to simulate

int no_days;

int day;

cout << "Enter initial amount of green crud:

";

cin >> newCrud;

cout << "Enter number of days to simulate:

";

cin >> no_days;

for (day = 10; day<=no_days; day++)

{

if (day % 10 == 0)

{

next = newCrud + init;

}

newCrud = init;

init = next;

}

if (no_days < 5)

cout << "\nCrud reproduce only after 5 days

minimum.Hence the current amount is "

<< newCrud << " pounds.";

else

cout << "On day " << no_days << " you have "

<< init

<< " pounds of green crud." << endl;

cout << "\nWould you like to continue? (y or

n): ";

cin >> ans;

}

return 0;

}

Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

Would you like to continue? (y or n): y

Enter initial amount of green crud: 5

Enter number of days to simulate: 225

On day 225 you have 10485760 pounds of green crud.

Using do while loop

Program:

#include <iostream>

using namespace std;

int main()

{

char ans;

do

{

//Initial crud size

int init = 0;

int newCrud;

int next=0;

//Number of days to simulate

int no_days;

int day;

cout << "Enter initial amount of green crud: ";

cin >> newCrud;

cout << "Enter number of days to simulate: ";

cin >> no_days;

for (day = 10; day<=no_days; day++)

{

if (day % 10 == 0)

{

next = newCrud + init;

}

newCrud = init;

init = next;

}

if (no_days < 5)

cout << "\nCrud reproduce only after 5 days

minimum.Hence the current amount is "

<< newCrud << " pounds.";

else

cout << "On day " << no_days << " you have " <<

init << " pounds of green crud." << endl;

cout << "\nWould you like to continue? (y or n):

";

cin >> ans;

} while (ans == 'Y' || ans == 'y');

return 0;

}

Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

Would you like to continue? (y or n): y

Enter initial amount of green crud: 5

Enter number of days to simulate: 225

On day 225 you have 10485760 pounds of green crud.

7 0

3 years ago

Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate

Monica [59]

Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

Explanation:

Here we have the heat Q given as follows;

Q = 15 × 1075 × (1100 - $t_{A2}$ ) = 10 × 1075 × (850 - 300) = 5912500 J

∴ 1100 - $t_{A2}$ = 1100/3

$t_{A2}$ = 733.33 K

$\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}$

Where

$\Delta \bar{t}_{a}$ = Arithmetic mean temperature difference

$t_{A_{1}$ = Inlet temperature of the gas = 1100 K

$t_{A_{2}$ = Outlet temperature of the gas = 733.33 K

$t_{B_{1}$ = Inlet temperature of the air = 300 K

$t_{B_{2}$ = Outlet temperature of the air = 850 K

Hence, plugging in the values, we have;

$\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K$

Hence, from;

$\dot{Q} = UA\Delta \bar{t}_{a}$ , we have

5912500 = 90 × A × 341.67

$A = \frac{5912500 }{90 \times 341.67} = 192.3 \, m^2$

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².

4 0

3 years ago

Harmonic excitation of motion is represent as

Gennadij [26K]

Harmonic excitation refers to a sinusoidal external force of a certain frequency applied to a system. ... Resonance occurs when the external excitation has the same frequency as the natural frequency of the system. It leads to large displacements and can cause a system to exceed its elastic range and fail structurally.

6 0

3 years ago