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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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An electric field of 2.09 kV/m and a magnetic field of 0.358 T act on a moving electron to produce no net force. If the fields a
My name is Ann [436]

Answer:

The velocity is  v =  5838 \ m/s

Explanation:

From the question we are told that

   The electric field is E  =  2.09 kV/m =  2.09 *10^{3} \ V/m

    The magnetic field is  B  =  0.358 \ T

     

Generally the force experienced by the electron due to the magnetic field is

         F_m  =  qvB

Generally the force experienced by the electron due to the electric  field is

       F_e =  qE

Since from the question the net force is zero  then

     F_e =  F_m

=>    v =  \frac{E}{B}

Substituting values

      v =  \frac{2.09*10^{3}}{0.358 }

    v =  5838 \ m/s

     

8 0
2 years ago
A uniform horizontal bar of mass m1 and length L is supported by two identical massless strings. String A Both strings are verti
NeX [460]

Answer:

a)  T_A = \frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} ) ,  b) T_B = g [m₂ ( \frac{x}{d} -1) + m₁ ( \frac{L}{ 2d} -1) ]

c)  x = d - \frac{m_1}{m_2} \  \frac{L}{2d},  d)  m₂ = m₁  ( \frac{ L}{2d} -1)

Explanation:

After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act

a) The tension of string A is requested

The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive

      ∑ τ = 0

      T_A d - W₂ x -W₁ L/2 = 0

      T_A = \frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )

b) the tension in string B

we write the expression of the translational equilibrium

       ∑ F = 0

       T_A - W₂ - W₁ - T_B = 0

       T_B = T_A -W₂ - W₁

       T_ B =   \frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )  - g m₂ - g m₁

       T_B = g [m₂ ( \frac{x}{d} -1) + m₁ ( \frac{L}{ 2d} -1) ]

c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system

         T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0

at the point that begins to rotate T_B = 0

          g m₂ (d -x) -  g m₁  (0.5 L -d) + 0 = 0

          m₂ (d-x) = m₁ (0.5 L- d)

          m₂ x = m₂ d - m₁ (0.5 L- d)

          x = d - \frac{m_1}{m_2} \  \frac{L}{2d}

 

d) The mass of the block for which it is always in equilibrium

this is the mass for which x = 0

           0 = d - \frac{m_1}{m_2} \  \frac{L}{2d}

         \frac{m_1}{m_2} \ (0.5L -d) = d

          \frac{m_1}{m_2} = \frac{ d}{0.5L-d}

          m₂ = m₁  \frac{0.5 L -d}{d}

          m₂ = m₁  ( \frac{ L}{2d} -1)

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