Which of the following would survive in a semi-arid to arid region of the western United States? a. short grasses and cacti b. tall deciduous trees c. tall coniferous trees d. tropical vegetation

Letter C is the correct answer.

The share-tenancy was a better option if compared to the sharecropping system. In the system of share-tenancy, farmers had more control over their crops than with sharecropping. Farmers had the freedom to operate tha land using the tecniques they wished and giving a percentage of the harvest as payment to the landowner. However, in the sharecropping system, farmers didn't have the same freedom and landowners would lie about expenses to keep the sharecroppers in debt.  

Answer:

C - Farmers had more control over their crops than sharecropping.

Explanation:

In tenant farming, the famers controlled what they planted and supplied everything themselves, giving the landowners a portion of the crops as payments - some tenent farmers even owned a small portion of their own land.

Sharecroppers were more at the landowners discretion with the landowners making the majority of the decisions and the sharecroppers only working the land.

A) See ray diagram in attachment (-6.0 cm)

By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f = -10 cm is the focal length (negative for a diverging lens)

p = 15 cm is the distance of the object from the lens

Solving for q,

\frac{1}{q}=\frac{1}{-10 cm}-\frac{1}{15 cm}=-0.167 cm^{-1}

q=\frac{1}{-0.167 cm^{-1}}=-6.0 cm

B) The image is upright

As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:

h_i = - h_o \frac{q}{p}

where h_i, h_o are the size of the image and of the object, respectively.

Since q < 0 and p > o, we have that h_i >0, which means that the image is upright.

C) The image is virtual

As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.

This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual

The image distance from the lens is 6 cm.

The image formed by the diverging lens is upright and virtual.

Image distance

The image distance is calculated by using the following lens formula;

\frac{1}{v} = \frac{1}{f} - \frac{1}{u}

where;

• v is the image distance
• f is the focal length
• u is the object distance

\frac{1}{v} = \frac{-1}{10} - \frac{1}{15} \\\\\frac{1}{v} = \frac{-15 - 10}{150} \\\\\frac{1}{v} = \frac{-25}{150} \\\\v = -6 \ cm

The basic properties of image formed by a diverging lens

• Image is reduce
• Image is upright
• Image is virtual.

Answer:

Mobilize

Explanation:

Mobilize is the resource management task that deploys or activates personnel and resources.

In the event of being notified through established channels, mobilization and of personnel and other resources begin.

This are done by a human resource manager who are responsible for deployment of labor and resources during incidents.

1. 16.3 A

To calculate the average rms current, we have to calculate the average rms current first. The peak voltage of the power line is Vp = 120 V, so the rms voltage is

V_{rms}=\frac{V_p}{\sqrt{2}}=\frac{120 V}{\sqrt{2}}=84.9 V

The family uses 1000 kWh of energy per month. Since

1 kW = 1000 W

1 h = 3600 s

This energy corresponds to

E=1000 kWh \cdot (1000 W/kW)\cdot (3600 s/h)=3.6\cdot 10^9 Ws=3.6\cdot 10^9 J

The time interval considered is 1 month, which corresponds to

t=1 m = (30 d)(24 h)(60 min)(60 s)=2.6\cdot 10^6 s

And the power is the ratio between the energy used and the time taken:

P=\frac{E}{t}=\frac{3.6\cdot 10^9 J}{2.6\cdot 10^6 s}=1385 W

And now we can find the average rms current by using the relationship between power, voltage and current:

I_{rms}=\frac{P}{V_{rms}}=\frac{1385 W}{84.9 V}=16.3 A

2. 5.2 \Omega

The average resistance of the household can be calculated by using Ohm's law. Using the value found for the rms voltage and the rms current, we find

R=\frac{V_{rms}}{I_{rms}}=\frac{84.9 V}{16.3 A}=5.2 \Omega

• The average rms current in the power line to the house is 12A
• The average resistance of a household is 10Ω.

Given the data in the question;

Electricity used; E = 1000kWh = 36*10^8J

Time; t = 1 month = 2.592 * 10^6s

Voltage; V = 120V

Calculate the average rms current to the house.

First we determine the power

Power is the amount of energy transferred per unit time:

Power = \frac{E}{t}

So we substitute in our values

Power = \frac{36*10^8J}{2.592*10^6s}\\\\Power = 1.388 * 10^3 J/s \\\\Power = 1.388 * 10^3 W

Next we Calculate the Current

From Ohms Law:

P = I * V

We substitute in our values

1.388*10^3W = I \ *\ 120V \\\\I = \frac{1.388*10^3W}{120V} \\\\I = 11.57 W/V\\\\I = 12A

Therefore, the average rms current in the power line to the house is 12A

The resistance of a household

From Ohm's Law:

R = \frac{V}{I}

We substitute in our values

R = \frac{120V}{12A}\\\\R = 10ohms

Therefore, The average resistance of a household is 10Ω.