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mylen [45]
3 years ago
12

Wildlife biologists are attempting to monitor the population size of the insect pest known as the gypsy moth (Lymantria dispar)

during the summer months. To estimate the total population size of these insects living in a 50 square kilometer forest, biologists have sampled random trees, counting the number of eggs and larvae on each randomly selected tree. Which of these factors MOST directly impacts the accuracy of this estimate?
A)
the number of trees sampled


B)
the presence of gypsy moth predators


C)
the overall health of the gypsy moth population


D)
the weather conditions at the time of the sampling
Physics
2 answers:
Elden [556K]3 years ago
5 0

Answer: Option A: The number of trees sampled.

The accuracy can be understood as how close is the measured value to the true value.   The aim is to monitor the population size of the insect pest in a 50 square kilometer. Random trees are selected, and number of eggs and larvae are counted. So, the measured value would be closer to actual value when the number of trees sampled are increased. More the number of trees sampled, less would be the chances of error and the accuracy of the estimate would increase.

Vlad [161]3 years ago
4 0

Answer:

a.) the number of tress sampled

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maw [93]

Answer:

Option (e)

Explanation:

A = 45 cm^2 = 0.0045 m^2, d = 0.080 mm = 0.080 x 10^-3 m,

Energy density = 100 J/m

Let Q be the charge on the plates.

Energy density = 1/2 x ε0 x E^2

100 = 0.5 x 8.854 x 10^-12 x E^2

E = 4.75 x 10^6 V/m

V = E x d

V = 4.75 x 10^6 x 0.080 x 10^-3 = 380.22 V

C = ε0 A / d

C = 8.854 x 10^-12 x 45 x 10^-4 / (0.080 x 10^-3) = 4.98 x 10^-10 F

Q = C x V = 4.98 x 10^-10 x 380.22 = 1.9 x 10^-7 C

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3 years ago
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Explanation:

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2 years ago
Positive Charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point charge q is located on the positiv
deff fn [24]

Answer:

 electric field E = - k Q (1 /r(r-a)), force    F = - k Q qo / r (r-a) and force for r>>a    F ≈ - k Q qo / r²

Explanation:

You are asked to find the electric field of a continuous charge distribution, so we must use the equation

       

           E = k ∫dp /r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C², r is the distance between the load distribution and the test charge, in this case everything is on the X axis.

We must find the charge differential (dq), let's use that uniformly distributed and create a linear charge density

          λ = q / x

As it is constant, we can write it based on differentials

         λ = dq / dx

         dq = λ dx

We already have all the terms, let's  integrate enter its limits, lower the distance from the left end of the distribution to the test charge (x = r) and the upper limit that is the distance from the left end of distribution to the test load ( x = r - a) where r> a

         E = k ∫ λ dx / x²

         E = k la (- 1 / x)

Let's get the negative sign from the parentheses

         E = - k λ (1 / x)

         E = - k λ (1 /(r-a)  -1 /r) = - k λ [a / r (r-a)]

Let's change the charge density with the value of the total charge λ = Q / a

         E = - k Q/a  [a / r (r-a)]

         E = - k Q (1 /r(r-a))

b) We calculate the force.  

         F = E qo

         F = - k Q qo / r (r-a)

c) the force for charge porbe very far r >> a. In this case we can take r from the parentheses and neglect (a/r)

         F = - k Qqo / r² (1 -  a/r)

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3 years ago
Calculate the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m
otez555 [7]

Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

Explanation:

Given: Mass = 5 kg

Spring constant = 6 N/m

Formula used to calculate period is as follows.

T = 2 \pi \sqrt\frac{m}{k}

where,

T = period

m = mass

k = spring constant

Substitute the values into above formula as follows.

T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s

Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

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