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denpristay [2]
4 years ago
15

What animals were delisted in july 2019?

Physics
1 answer:
andrezito [222]4 years ago
5 0

The animals which were delisted from the endangered species in July 2019 is Grey Wolf

Answer: Grey wolf

<u>Explanation:</u>

The ultimate goal of the Endangered Species Act was to preserve the species that are in the verge of extinction. The species are conserved and taken care of till they can be left free in the wild for it to survive of it's own.If the species has crossed all the recovery goals then they are delisted.

The grey wolf in July 2019 was delisted from the endangered species list. Now it is calculated that there are 6000 wolves that are found. But yet the grey wolf has not returned back to their livelihood and also to their sutable habitat.

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a student standing in an elevator at rest notices that her pendulum has a frequency of oscillation of 0.5 Hz. a. what is the len
EastWind [94]

Answer:

L = 0.99 m = 99 cm

Explanation:

The period is the reciprocal of the frequency.

T = 1/0.5 = 2.0 s

T = 2π√(L/g)

L = g(T/2π)²

L = 9.8(2.0/2π)² = 0.99 m

If the system accelerates upward, it will cause the apparent gravity to increase. This will require a longer pendulum to keep the same period, or shorten the period if the length remains the same. This shows up in the equation where the product of gravity and the square of the period must remain constant for the length to remain constant.

5 0
3 years ago
An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity σ1 = 0.51 μC/m2. Another infini
NNADVOKAT [17]

Answer:

 E_total = 5.8 10⁴ N /C

Explanation:

In this problem they ask to find the electric field at two points, the electric field is a vector magnitude, so we can find the field for each charged shoah and add them vectorally at the point of interest.

To find the electric field of a charged conductive sheet, we can use the Gauss law,

        Ф = E. d S = q_{int} / ε₀

Let us use as a Gaussian surface a small cylinder, with the base parallel to the sheet, the electric field between the sheet and the normal one next to the cylinder has 90º, so its scalar product is zero, the electric field between the sheet and the base has An Angle of 0º, therefore the scalar product is reduced to the algebraic product.

Let's look for the electric field for plate 1

The total flow is the same for each face, as there are two sides of the cylinder

       2E A = q_{int} /ε₀

For the internal load we use the concept of surface density

      σ = q_{int1} / A

      q_{int1} = σ₁ A

Let's replace

       2E A = σ₁ A /ε₀

        E₁ = σ₁ / 2ε₀

For the other plate we have a field with a similar expression, but of negative sign

       E₂ = -σ₂ / 2ε₀

The total field is,

        E_total = σ₁ / 2ε₀ + σ₂ / 2ε₀

       E_total = (σ₁ + σ₂) / 2ε₀

Let us apply this expression to our case, when placing a sheet without electric charge, a charge is induced for each sheet, the plate 1 that has a positive charge the electric field is protruding to the right and the plate 2 that has a negative charge creates a incoming field, to the right, as the two fields have the same address add

           The conductive sheet in the middle pate undergoes an induced load that is created by the other two plates, but because the conductive plate the charges are mobile and are replaced.

       E_total = (0.51 +0.52) 10⁻⁶ / 2 8.85 10⁻¹²

       E_total = 5.8 10⁴ N /C

Note that the field is independent of the distance between the plates

4 0
3 years ago
Lol somebody tell me the answer ill also give 5 stars
sdas [7]
Hi C is the Answer I believe
3 0
3 years ago
A squirrel pulls a 8.7 kg bag of acorns along the ground with a constant force of 80 N. If the force of friction opposing this s
Annette [7]

Answer:

a=5.51\ m/s^2

Explanation:

Given that,

The mass of a bag of acorns, m = 8.5 kg

The force acting on the bag = 80 N

The force of friction = 32 N

We need to find the acceleration of the bag of acorns.

Net force acting on it is = 80 N - 32 N

= 48 N

Let a be the acceleration of the bag of acorns. So,

F = ma

a=\dfrac{F}{m}\\\\a=\dfrac{48}{8.7}\\\\a=5.51\ m/s^2

So, the acceleration of the bag of acorns is 5.51\ m/s^2.

5 0
3 years ago
What acceleration will result when a 12 N net force applied to a 3 kg object?​
geniusboy [140]

Answer:

4m/s

Explanation:

due to newtons second law of motion

the accelerations that result when a 12-N net force is applied to a 3-kg object. A 3-kg object experiences an acceleration of 4 m/s/s.

HOPE THIS HELPS PLEASE MARK AS BRAINLIEST:)

8 0
4 years ago
Read 2 more answers
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