The machine which turns in a power station

Alcohol occurs when the desire to feel good and have the pleasurable feelings associated with alcohol or other drugs drives desi

IRISSAK [1]

Ans: Alcohol addiction occurs when the desire to feel good and have the pleasurable feelings associated with alcohol or other drug drives desire to continue using the substances.
This is because when a person consumes alcohol on a daily basis, that person then becomes dependent on alcohol and then in every mood of happiness or anxiety, he/she has a strong desire of consuming alcohol and feel good after consuming it.

7 0

3 years ago

What are simple machines please?

lubasha [3.4K]

A basic mechanical device that is used for applying a force.

Examples are: inclined plane, lever, wedge, wheel and axle, pulley, and screw...Hope this helps, have a BLESSED and wonderful day!

5 0

3 years ago

Read 2 more answers

What are the chances that radio transmissions from Earth or messages sent on distant space probes will ever be received by livin

spin [16.1K]

Answer:

Very low

Explanation:

It is fair to say to a reasonable extent that there are very low chances that radio transmissions from Earth or messages sent on distant space probes will ever be received by living beings .

Bear in mind that for years some scientists have believed without any substantial evidence that there are other living beings in distant space.

4 0

3 years ago

Argon makes up 0.93% by volume of air. Calculate its solubility (in mol/L) in water at 20°C and 1.0 atm. The Henry's law constan

vladimir2022 [97]

Answer:

S= 1.40x10⁻⁵mol/L

Explanation:

The Henry's Law is given by the next expression:

$S = k_{H} \cdot p$ (1)

where S: is the solubility or concentration of Ar in water, $k_{H}$ : is Henry's law constant and p: is the pressure of the Ar

Since the argon is 0.93%, we need to multiply the equation (1) by this percent:

$S = 1.5 \cdot 10^{-3} \frac{mol}{L\cdot atm} \cdot 1.0atm \cdot \frac{0.93}{100} = 1.40 \cdot 10^{-5} \frac{mol}{L}$

Therefore, the argon solubility in water is 1.40x10⁻⁵mol/L.

Have a nice day!

8 0

3 years ago

An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su

s2008m [1.1K]

Answer: A. 23.59 minutes.

B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

Latent heat of fusion is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

Specific heat capacity of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, given that:

mass of ice, m= 0.555 kg

temperature of ice, T= -16.6°C

rate of heat transfer, $q=820 J.min^{-1}$

specific heat of ice, $c_{i}= 2100 J.kg^{-1}.K^{-1}$

latent heat of fusion of ice, $L_{i}=334\times10^{3}J.kg^{-1}$

Asked:

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

We have the formula:

$Q=mc\Delta T$

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

$Q= 0.555\times2100\times16.6$

$Q= 19347.3 J$

Now, we require 19347.3 joules of heat to bring the ice to 0°C and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

$t=\frac{Q}{q}$

$=\frac{19347.3}{820}$

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

Now comes the concept of Latent heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

We have the equation: latent heat, $Q_{L}= mL_{i}$