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Ksivusya [100]
3 years ago
9

Based on what you have read provide two positives and two negatives the Suarez family would face if they switch to organic farmi

ng
Physics
1 answer:
arlik [135]3 years ago
6 0

Hi, you did not enter the text about the Suarez family, which makes it impossible for this question to be answered accurately. However, from the context of your question, it is possible to see that you want to know two positive and negative points that the Suarez family would observe when starting a planting of organic agriculture.

Among the negative points, we can say that organic agriculture has a high cost, as the techniques used in it require a greater number of resources, which can make not only the service more expensive, but also the product. A second negative point is that the techniques used in organic agriculture are more difficult than the techniques used in conventional agriculture, since all management of organic crops must be carried out without the application of chemical products.

Among the positive points, we can say that the growing number of consumers of organic products can increase the profit that the family will have from the sale of these products. In addition, the family will have healthier, less polluted and more productive land.

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In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
3 years ago
Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has a magnitude of 3.7m/s2.​ We want
lbvjy [14]

Explanation:

It is given that, Onur drops a basketball from a height of 10 m on Mars, where the acceleration due to gravity has a magnitude of 3.7 m/s².

The second equation of kinematics gives the relationship between the height reached and time taken by it.

Here, the ball is droped under the action of gravity. The value of acceleration due to gravity on Mars is positive.

We want to know how many seconds the basketball is in the air before it hits the ground. So, the formula is :

\Delta x=v_ot+\dfrac{1}{2}at^2

t is time taken by the ball to hit the ground

v_o is initial speed of the ball

So, the correct option is (A).

8 0
3 years ago
How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0
3 years ago
The melting point of a substance is the same as its _____ point.
kirza4 [7]

Answer:

Boiling point

Explanation:

5 0
2 years ago
Blake stands in a canoe in the middle of a lake. The canoe is stationary. Blake holds an anchor mass of 15 kg, then throws it we
Inessa05 [86]

The velocity of the canoe is  1.7 m/s.

<h3>What is momentum?</h3>

Momentum in physics is the products of mass and velocity. Now we have to find momentum with the formula; p = mv

a) Initial momentum = (15)8 m/s + 135 = 255 Kgms-1

b) Since momentum is conserved, the total momentum after throwing the anchor is still 255 Kgms-1

c) The final velocity of the boat is obtained from;

255 Kgms-1 = (15Kg + 135 Kg) v

v = 255 Kgms-1/(15Kg + 135 Kg)

v = 1.7 m/s

Learn more about momentum: brainly.com/question/904448

5 0
2 years ago
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