Motion maps for two objects, Y and Z, are shown.

Imagine that you work for a conservation organization. Your job is to make recommendations to restore an area of a prairie that

ella [17]

Answer:

The damaged prairie area can be restored through site preparation, replanting and putting species back on the landscape.

Explanation:

Prairie- this is a type of habitat with mostly grasses. It also consists of trees, occasional shrubs and flowering plants. This can be found around the world and is ecologically important for providing habitats for native animals and insects. It is perfectly adapted to the climate and requires minimal maintenance.

Once a prairie is damaged, it becomes an essential move to restore it immediately. This can be done through different human activities such as the following:

1. Site preparation- before doing anything else, it would be best to prepare the prairie site by removing debris from the fire. This will make everything easier once you start planting the crops or plants.

2. Replanting- this is probably the most essential move you can do. It would be best to plant high quality food plants that would provide nectar, pollen and seeds throughout the season. This will entice many birds and butterflies into the area. It will also make the native plants abundant which will, in turn, increase ecological diversity.

3. Putting species back on the landscape- when a wildfire occurs, it is a common sense that species also run away from their home (habitat). In order to create that natural habitat again, it would be best to put the species back to where they belong so they can restart their lives. This will resume the natural food webs and nutrient cycles.

7 0

3 years ago

An electric motor supplies 2,500 Watts of power in a 220 V circuit. What is the resistance of the motor? Be sure to include the

s344n2d4d5 [400]

Answer:

R = 19.36 ohms

Explanation:

Given that,

The power of an electric motor, P = 2500 Watts

The voltage of the circuit, V = 220 V

We need to find the resistance of the motor. We can find it as follows :

$P=\dfrac{V^2}{R}\\\\R=\dfrac{V^2}{P}\\\\R=\dfrac{(220)^2}{2500}\\R=19.36\ \Omega$

So, the resistance of the motor is 19.36 ohms.

4 0

3 years ago

Which is true of the buoyant force?

victus00 [196]

It acts in the upward direction.

8 0

3 years ago

Un contenedor de 1800 N está en reposo sobre un plano inclinado a 28°, el coeficiente de fricción entre el contenedor y el plano

babunello [35]

Answer:

F = 1480.77N

Explanation:

In order to calculate the required force to push the container with a constant velocity, you take into account the the sum of force on the container is equal to zero. Furthermore, you have for an incline the following sum of forces:

$F-Wsin\alpha-F_r=0\\\\F-Wsin\alpha-N\mu cos\alpha=0\\\\F-Wsin\alpha-W\mu cos\alpha=0$ (1)

F: required force = ?

W: weight of the container = 1800N

N: normal force = weigth

α: angle of the incline = 28°

g: gravitational acceleration = 9.8m/s^2

μ: coefficient of friction = 0.4

You solve the equation (1) for F and replace the values of the other parameters:

$F=W(sin\alpha+\mu cos\alpha)\\\\F=(1800N)(sin28\°+(0.4)cos28\°)=1480.77N$

The required force to push the container for the incline with a constant velocity is 1480.77N

8 0

3 years ago

Suppose Gabor, a scuba diver, is at a depth of 15m. Assume that: The air pressure in his air tract is the same as the net water

s2008m [1.1K]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

a

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at

the surface is $\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = 2.5$

b

The number of moles of gas that must be released is $n= 0.3538\ mols$

Explanation:

We are told from the question that the pressure at the surface is 1 atm and for each depth of 10m below the surface the pressure increase by 1 atm

This means that the pressure at the depth of the surface would be

$P_d = [\frac{15m}{10m} ] (1 atm) + 1 atm$

$= 2.5 atm$

The ideal gas equation is mathematically represented as

$PV = nRT$

Where P is pressure at the surface

V is the volume

R is the gas constant = 8.314 J/mol. K

making n the subject we have

$n = \frac{PV}{RT}$

Considering at the surface of the water the number of moles at the surface would be

$n_s = \frac{P_sV}{RT}$

Substituting $1 atm = 101325 N/m^2$ for $P_s$ , $6L = 6*10^{-3}m^3$ for volume , 8.314 J/mol. K for R , (37° +273) K for T into the equation

$n_s = \frac{(1atm)(6*10^{-3} m^3)}{(8.314J/mol \cdot K)(37 +273)K}$

$= 0.2359 mol$

To obtain the number of moles at the depth of the water we use

$n_d = \frac{P_d V}{RT}$

Where $P_d \ and \ n_d \$ are pressure and no of moles at the depth of the water

Substituting values we have

$n_d = \frac{(2.5)(101325 N/m^2)(6*10^{-3}m^3)}{(8.314 J/mol \cdot K)(37 + 273)K}$

$= 0.5897 mol$

Now to obtain the number of moles released we have

$n = n_d - n_s$

$= 0.5897mol - 0.2359mol$

$=0.3538 \ mol$

The molar concentration at the surface of water is

$[\frac{n}{V} ]_{surface} = \frac{0.2359mol}{6*10^-3m^3}$

$=39.31mol/m^3$

The molar concentration at the depth of water is

$[\frac{n}{V} ]_{15m} = \frac{0.5897}{6*10^{-3}}$

$= 98.28 mol/m^3$

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at the surface is

$\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = \frac{98.28}{39.31} =2.5$

6 0

3 years ago