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3 years ago

Why do farmers need to use strategies like crop rotation?

Physics

2 answers:

Mrrafil [7]3 years ago

6 0

Answer:

Crops drain the soil of nutrients

Explanation:

i took the test:) yw

alexdok [17]3 years ago

4 0

Answer:

Crops drain the soil of nutrients

Explanation:

Trust me fam

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In a stadium, fans stand up and sit down to produce a wave across the stadium. This type of wave where the material travels perp

puteri [66]

Correct option B

Transverse waves are those waves whose particles vibrate perpendicular to the direction of wave.

Hope This Helps You!

3 0

3 years ago

Read 2 more answers

Can you please answer these ASAP!

SVEN [57.7K]

Answer:

Let's explain this briefly.

Suppose that we have a piece of ice (this is, solid water) now we give energy to the piece of ice, so the temperature of the ice increases. There is a point where the piece of ice will start a change of phase, at this point the temperature of the ice stops increasing because all the energy we give to the ice is used in the change of phase.

Once we have a complete change of phase, the temperature can increase again, and now we will have liquid water.

If we keep increasing the temperature we will see this happen again, when we have the transition from liquid to gas.

(and a similar thing happen when we have a material in a given phase and we remove heat from the material).

In the images we can see the different changes of phase of water.

1) In the first image we can see the circle in a part where the temperature is constant, so the temperature does not change in this part, which means that there is a change of phase happening.

2) Here we have the circle in a diagonal line, so here the temperature is changing, meaning that we have an increase of temperature in this region.

3) Here we want to know what the x-axis represents, this should rerpesent the energy that is being given to the material (so in some parts we see that the temperature increases and in other parts we see that the material changes of phase)

Then here the correct option is heat over time.

4) The freezing point is the temperature in which the change of phase from liquid to solid happens (or solid to liquid).

In the graph we can see that this change of phase happens at the temperature T = -210°C

Then the correct option is -210°C (The last option)

4 0

3 years ago

The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is

stealth61 [152]

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

$\rho_m = 846lb/ft^3$

$g = 32.17405ft/s^2$

$h_1 = 1in = \frac{1}{12} ft$

For the air the defined properties would be

$\rho_a = 0.0075lb/ft^3$

$g = 32.17405ft/s^2$

$h_2 = ?$

We have for equilibrium that

$\text{Pressure change in Air}=\text{Pressure change in Mercury}$

$\rho_m g h_1 = \rho_a g h_2$

Replacing,

$(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)$

Rearranging to find $h_2$

$h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}$

$h = 9400ft$

Therefore the elevation of the mountain top is 9400ft

7 0

3 years ago

Which will produce a magnetic field?

Yanka [14]

Answer:

a magnet

3 0

3 years ago

ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami

yuradex [85]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

Explanation :

The given balanced chemical reaction is,

$2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{product}$

$\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]$

$\Delta H^o=62.2kJ=62200J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{product}$

$\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]$

$\Delta S^o=132.67J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 298 K.

$\Delta G^o=(62200J)-(298K\times 132.67J/K)$

$\Delta G^o=22664.34J=22.66kJ$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

6 0

3 years ago