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abruzzese [7]
4 years ago
11

1. Butterflies usually are identified by the colors and patterns on their wings. Butterflies have four wings. The two wings near

the butterfly's head are "forewings while
ead are forewings while
the wings near the butterfly's tail are "hindwings. You can use the following dichotomous key to identify four common butterfly species in North America.
i
ii.
Hindwings pointed
Go to 2
Hindwings rounded ..................... Goto 3
Wings mainly yellow ............... Papilio glaucus
Wings mainly black .............. Papilio polyxenes
Wings orange with black piping...Danaus plexippus
Wings yellow with black edges..... Colias philodice
ii.
i.
i.
Based on the key, what speries of butterfly is shown below?
A Papilio glaucus
Physics
1 answer:
tia_tia [17]4 years ago
7 0

Answer:

Can't answer without a picture

Explanation:

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Transfer of heat through conduction can occur in which of the following?
lianna [129]

Answer:d I think

Explanation:

8 0
3 years ago
Read 2 more answers
List the steps involved in the nitrogen cycle
Nesterboy [21]
Nitrogen fixation (N2 to NH3/ NH4+ or NO3-)
Nitrification (NH3 to NO3-)
Assimilation (Incorporation of NH3 and NO3- into biological tissues)Ammonification (organic nitrogen compounds to NH3)
<span>Denitrification(NO3- to N2)

its on google</span>
4 0
4 years ago
The drawing shows a tire of radius R on a moving car
masha68 [24]

Answer:

 H / R = 2/3

Explanation:

Let's work this problem with the concepts of energy conservation. Let's start with point P, which we work as a particle.

Initial. Lowest point

          Em₀ = K = 1/2 m v²

Final. In the sought height

         Em_{f} = U = mg h

Energy is conserved

        Em₀ =  Em_{f}

        ½ m v² = m g h

        v² = 2 gh

Now let's work with the tire that is a cylinder with the axis of rotation in its center of mass

Initial. Lower

        Em₀ = K = ½ I w²

Final. Heights sought

        Emf = U = m g R

        Em₀ =  Em_{f}

       ½ I w² = m g R

The moment of inertial of a cylinder is

       I = I_{cm} + ½ m R²

      I= ½ I_{cm}  + ½ m R²

Linear and rotational speed are related

       v = w / R

       w = v / R

We replace

      ½ I_{cm}  w² + ½ m R² w²  = m g R

moment of inertia of the center of mass      

      I_{cm}  = ½ m R²

     ½ ½ m R² (v²/R²) + ½ m v² = m gR

    m v² ( ¼ + ½ ) = m g R

   

       v² = 4/3 g R

As they indicate that the linear velocity of the two points is equal, we equate the two equations

        2 g H = 4/3 g R

         H / R = 2/3

7 0
3 years ago
What is the energy of the photon that could cause (a) an electronic transition from then= 4 state to the n 5 state of hydrogen a
Fiesta28 [93]

Answer:

a) E_photon =0.306 eV

b) E_photon =0.166 eV

Explanation:

The energy of the photon (E) for n^th orbit of the hydrogen atom is given as:

E_photon = E_o(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})

where,

E_o = 13.6 eV

n = orbit

a) Now for the transition from n = 4 to n = 5

E_photon =13.6(\frac{1}{4^{2}}-\frac{1}{5^{2}})

E_photon =0.306 eV

b) Now for the transition from n = 5 to n = 6

E_photon =13.6(\frac{1}{5^{2}}-\frac{1}{6^{2}})

E_photon =0.166 eV

7 0
3 years ago
Two long, parallel wires each carry the same current I, but the two currents are anti-parallel. The two wires are a distance r a
Georgia [21]

Answer:

8 x 10⁻⁷ x  I / r

Explanation:

Two parallel long wires are carrying current I . Let the direction be towards the right in the farthest and towards the left in the nearest. Magnetic field due to current I  at a  distance d is given by the expression

B = μ₀ 2 I / 4π d

I the present case distance d = r/2  

Magnetic field due to one wire at point d = r/2  is

B₁ = μ₀ 2 I / (4π r / 2 )

= 10⁻⁷ x 4I / r

Magnetic field due to the other wire at point d = r/2  is

B₂ = μ₀ 2 I / (4π r / 2 )

= 10⁻⁷ x 4I / r

Direction of magnetic field due to both the wires at the mid  point P will be same . It will be in downward direction in the given scenario

So total magnetic field

B = B₁ + B₂

= 2 x  10⁻⁷ x 4I / r

= 8 x 10⁻⁷ x  I / r

6 0
3 years ago
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