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3 years ago

What animals were delisted in july 2019?

Physics

1 answer:

andrezito [222]3 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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The stars, Rigel and Betelgeuse, are both found in the constellation, Orion. Rigel is a blue supergiant, and Betelgeuse is a red

sattari [20]

Answer: a

Explanation: The color of a star is linked to its surface temperature. The hotter the star, the shorter the wavelength of light it will emit. The hottest ones are blue or blue-white, which are shorter wavelengths of light. Cooler ones are red or red-brown, which are longer wavelengths.

6 0

3 years ago

In the context of energy transfers with hot and cold reservoirs, the sign convention is that _______________.

Likurg_2 [28]

Answer:

B. QC > 0; QH < 0

Explanation:

Given that there are two reservoir of energy.

Sign convention for heat and work :

1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

From hot reservoir heat is going out that is why it is taken as negative

$Q_H$

From cold reservoir heat is coming inside the reservoir that is why it is taken as positive

$Q_C>0$

That is why the answer will be

$Q_H$ , $Q_C>0$

8 0

3 years ago

Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w

Tatiana [17]

Answer:

The average velocity is

$266\frac{m}{s},274\frac{m}{s}$ and $117\frac{m}{s}$ respectively.

Explanation:

Let's start writing the vertical position equation :

$L(t)=2t^{3}+t^{2}-5t+1$

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

$V_{avg}=\frac{Displacement}{Time}$ = Δx / Δt = $\frac{x2-x1}{t2-t1}$

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

$t2-t1=8s-5s=3s$

For the position variation we use the vertical position equation :

$x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m$

$x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m$

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

$\frac{798m}{3s}=266\frac{m}{s}$

For the second time interval :

t1 = 4 s → t2 = 9 s

$x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m$

$x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m$

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

$\frac{1370m}{5s}=274\frac{m}{s}$

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

$x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m$

$x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m$

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

$\frac{702m}{6s}=117\frac{m}{s}$

5 0

3 years ago

What happens when two minerals have different arrangements of Atoms

zaharov [31]

The overall arrangements of the atoms produce crystals

5 0

2 years ago

A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00

vivado [14]

Answer:

7.5 m

Explanation:

$v$ = initial speed of the ball = 8 m/s

$\theta$ = angle of launch = 40° deg

Consider the motion along the vertical direction :

$v_{oy}$ = initial velocity along vertical direction = $v Sin\theta$ = 8 Sin40 = 5.14 m/s

$a_{y}$ = acceleration along vertical direction = - 9.8 m/s²

$t$ = time of travel

$y$ = vertical displacement = - 1 m

Using the kinematics equation

$y = v_{oy}t + (0.5)a_{y}t^{2}$

$- 1 = (5.14)t + (0.5)(- 9.8)t^{2$

$t$ = 1.22 sec

Consider the motion along the horizontal direction :

$v_{ox}$ = initial velocity along horizontal direction = $v Sin\theta$ = 8 Cos40 = 6.13 m/s

$a_{x}$ = acceleration along vertical direction = 0 m/s²

$t$ = time of travel = 1.22 sec

$x$ = horizontal displacement = ?

Using the kinematics equation

$x = v_{ox}t + (0.5)a_{x}t^{2}$

$x = (6.13)(1.22) + (0.5)(0)(1.22)^{2$

$x$ = 7.5 m

4 0

2 years ago