W²-5w+14

The seers were of the opinion that_____ . *

AURORKA [14]

Answer:

✔️a healthy mind resides in a healthy body.

Explanation:

The seers were of the opinion that "a healthy mind resides in a healthy body."

Just like the English translation of a famous quotation from Thales, pre-Socratic Greek philosopher puts it "a sound mind in a sound body"; which tries to demonstrate the close connections that exists in bodily well-being and one's ability to enjoy life.

The seers were actually of the opinion that a healthy mind resides in a healthy body. It implies that there is connection between the body and the mind. When the body catches an illness, the mind and other parts of the body are affected. When our minds are not healthy, it affects the effective functioning of the body.

So, a healthy mind will definitely be found in a healthy body.

4 0

3 years ago

How much cornfield area would be required if you were to replace all the oil consumed in the United States with ethanol from cor

zaharov [31]

Answer:

2377.35 km

Explanation:

Given the following;

1. A cornfield is 1.5% efficient at converting radiant energy into stored chemical potential energy;

2. The conversion from corn to ethanol is 17% efficient;

3. A 1.2:1 ratio for farm equipment to energy production

4. A 50% growing season and,

5. 200 W/m2 solar insolation.

As per our assumptions,1.2/1 is the ratio for farm equipment to energy production,

So USA need around 45.45% (1/(1+1.2) replacement of fuel energy production which is nearly about = 0.4545*10^{20} J/year = \frac{0.4545*10^{20}}{365*24*3600}=1.44121*10^{12} J/sec

Growing season is only part of year ( Given = 50%),

Net efficiency = 1.5%*17%*50%=0.015*0.17*0.5=0.001275 = 0.1275%

Hence , Actual Energy replacement (Efficiency),

=\frac{1.44121*10^{12}}{0.001275} = 1.13*10^{15} J/sec=1.13*10^{15} W

As per assumption (5),

\because 200 W/m2 solar insolation arequired,

So USA required corn field area = 1.13*10^{15}/200 = 5.65*10^{12} m^{2}

Hence, length of each side of a square,

= (5.65*10^{12} )^{0.5} = 2377.35 km

4 0

3 years ago

Let r be a bank's interest rate in percent per year (APR). An initial amount of money P, also called as principal, will mature t

Liono4ka [1.6K]

Answer:

See explanation below.

Explanation:

For this case the program needs to take the inputs as P,r and n and the output would be as A and printed on the system. The code is:

# Inputs

P = float(input("Enter the present value : "))

r = float(input("Enter your APR : "))

n = float(input("Enter the number of years : ") )

# Output

A = P*(1 +(r/100))**n

print("The future values is:", A)

And the result obtained is:

Enter the present value : 1000

Enter your APR : 0.95

Enter the number of years : 5

The future values is: 1048.4111145526908

7 0

3 years ago

Nanotechnology is a scientific area that deals with making or changing things that are incredibly _______________. *

Dafna1 [17]

B tiny nano means small!!

3 0

3 years ago

Read 2 more answers

A rigid tank with a total volume of 0.05 m3 initially contains a two-phase liquid-vapor mixture of water at a pressure of 15 bar

Westkost [7]

Answer:

a) $m_{2} = 0.753\,kg$ , b) $Q_{in} = 2122.963\,kJ$

Explanation:

A rigid tank means a storage whose volume is constant. Process is entirely isobaric. Initial and final properties of water are included below:

State 1 - Gas-Vapor Mixture

$P = 1500\,kPa$

$T = 198.29^{\textdegree}C$

$\nu = 0.02726\,\frac{m^{3}}{kg}$

$u = 1192.94\,\frac{kJ}{kg}$

$h_{g} = 2791.0\,\frac{kJ}{kg}$

$x = 0.2$

State 2 - Gas-Vapor Mixture

$P = 1500\,kPa$

$T = 198.29^{\textdegree}C$

$\nu = 0.06643\,\frac{m^{3}}{kg}$

$u = 1718.12\,\frac{kJ}{kg}$

$h_{g} = 2791.0\,\frac{kJ}{kg}$

$x = 0.5$

The model for the rigid tank is created by using the First Law of Thermodynamics:

$Q_{in} - (m_{1}-m_{2})\cdot h_{g} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}$

Initial and final masses are:

$m_{1} = \frac{V_{1}}{\nu_{1}}$

$m_{1} = \frac{0.05\,m^{3}}{0.02726\,\frac{m^{3}}{kg} }$

$m_{1} = 1.834\,kg$

$m_{2} = \frac{V_{2}}{\nu_{2}}$

$m_{2} = \frac{0.05\,m^{3}}{0.06643\,\frac{m^{3}}{kg} }$

$m_{2} = 0.753\,kg$

a) The final mass within the tank is:

$m_{2} = 0.753\,kg$

b) The total amount of heat transfer is:

$Q_{in} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}+ (m_{1}-m_{2})\cdot h_{g}$

$Q_{in} = (0.753\,kg)\cdot (1718.12\,\frac{kJ}{kg} )- (1.834\,kg)\cdot (1192.94\,\frac{kJ}{kg} ) + (1.081\,kg)\cdot (2791.0\,\frac{kJ}{kg} )$

$Q_{in} = 2122.963\,kJ$

5 0

3 years ago