2 -An invasive species is a species that is not native to a specific location (an introduced species), and that has a tendency to spread to a degree believed to cause damage to the environment, human economy or human health.[2]
The term as most often used applies to introduced species that adversely affect the habitats and bioregions they invade economically, environmentally, or ecologically. Such species may be either plants or animals and may disrupt by dominating a region, wilderness areas, particular habitats, or wildland–urban interface land from loss of natural controls (such as predators or herbivores). This includes plant species labeled as exotic pest plants and invasive exotics growing in native plant communities.[3][4][5][6] The European Union defines "Invasive Alien Species" as those that are, firstly, outside their natural distribution area, and secondly, threaten biological diversity.[7][8] The term is also used by land managers, botanists, researchers, horticulturalists, conservationists, and the public for noxious weeds.[9]
The speed of the water in the wider part will be 1.194 m/sec. Speed is a time-based quantity. Its SI unit is m/sec.
<h3> What is speed?</h3>
Speed is defined as the rate of change of the distance or the height attained.
The given data in the problem is;
The initial diameter is,![\rm d_1 = 2.2 \ cm](https://tex.z-dn.net/?f=%5Crm%20d_1%20%3D%202.2%20%5C%20cm)
initial radius,
![r_1 = \frac{d_1}{2} \\\\ r_1 = \frac{2.2}{2} \\\\ r_1 = 1.1\ cm](https://tex.z-dn.net/?f=r_1%20%3D%20%5Cfrac%7Bd_1%7D%7B2%7D%20%5C%5C%5C%5C%20r_1%20%3D%20%5Cfrac%7B2.2%7D%7B2%7D%20%5C%5C%5C%5C%20r_1%20%3D%201.1%5C%20cm)
The initial crossection area;
![\rm A_1 = \pi r_1^2 \\\\ \rm A_1 = 3.14 \times (1.1\times 10^{-2})^2 \\\\ \rm A_1 =3.8 \times 10^{-4} \ m^2](https://tex.z-dn.net/?f=%5Crm%20A_1%20%3D%20%5Cpi%20r_1%5E2%20%5C%5C%5C%5C%20%5Crm%20A_1%20%3D%203.14%20%5Ctimes%20%20%281.1%5Ctimes%2010%5E%7B-2%7D%29%5E2%20%5C%5C%5C%5C%20%5Crm%20A_1%20%3D3.8%20%5Ctimes%2010%5E%7B-4%7D%20%5C%20m%5E2)
The final crossection area;
![\rm A_2 = \pi r_2^2 \\\\ \rm A_2 = 3.14 \times ( 2 \times 10^{-2})^2 \\\\ \rm A_2 = 12.56 \ m^2](https://tex.z-dn.net/?f=%5Crm%20A_2%20%3D%20%5Cpi%20r_2%5E2%20%5C%5C%5C%5C%20%5Crm%20A_2%20%3D%203.14%20%5Ctimes%20%28%202%20%5Ctimes%2010%5E%7B-2%7D%29%5E2%20%5C%5C%5C%5C%20%5Crm%20A_2%20%3D%2012.56%20%5C%20m%5E2)
The initial flow rate is;
R = density ×velocity ×area
![\rm R = \rho A V \\\\ 1.5 = 1000 \times V_1 \times 3.8 \times 10^{-4} \\\\ V_1 = 3.947 \ m/sec](https://tex.z-dn.net/?f=%5Crm%20R%20%3D%20%5Crho%20A%20V%20%5C%5C%5C%5C%201.5%20%3D%201000%20%5Ctimes%20V_1%20%5Ctimes%203.8%20%5Ctimes%2010%5E%7B-4%7D%20%5C%5C%5C%5C%20V_1%20%20%3D%203.947%20%5C%20m%2Fsec)
The speed of the water in the wider part will be;
From the continuity equation;
![\rm A_1 V_1 = A_2V_2 \\\\\ 3.8 \times 10^{-4} \times 3.947 = 12.56 \times 10^{-4} \times V_2 \\\\ V_2= 1.194 \ m/sec](https://tex.z-dn.net/?f=%5Crm%20A_1%20V_1%20%3D%20A_2V_2%20%20%5C%5C%5C%5C%5C%203.8%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctimes%203.947%20%3D%2012.56%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctimes%20V_2%20%5C%5C%5C%5C%20V_2%3D%201.194%20%5C%20m%2Fsec)
Hence, the speed of the water in the wider part will be 1.194 m/sec.
To learn more about the speed, refer to the link;
brainly.com/question/7359669
#SPJ1
Answer:
a is the answer to the question
The correct answer of the given question above would be option B. The statement that is not correct is that, a steady magnetic field produces a steady current. The rest of the statements are all correct. <span>An unchanging/static magnetic field (relative to a wire/circuit) induces zero current.</span>
Answer:
<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong):</em>
a) a×b = 34.27k
b) a·b = 128.43
c) (a + b)·b = 305.17
d) The component of a along the direction of b = 9.66
Explanation:
<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong)</em> we can proceed as follows:
a) The vectorial product, a×b is:
![a \times b = (8.6*9.5 - 5.1*9.3)k = 34.27k](https://tex.z-dn.net/?f=%20a%20%5Ctimes%20b%20%3D%20%288.6%2A9.5%20-%205.1%2A9.3%29k%20%3D%2034.27k%20)
b) The escalar product a·b is:
![a\cdot b = (8.6*9.3) + (5.1*9.5) = 128.43](https://tex.z-dn.net/?f=%20a%5Ccdot%20b%20%3D%20%288.6%2A9.3%29%20%2B%20%285.1%2A9.5%29%20%3D%20128.43%20)
c) <u>Asumming (a</u><u> </u><u>+ b)·b</u> <em>instead a+b·b</em> we have:
![(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17](https://tex.z-dn.net/?f=%28a%20%2B%20b%29%5Ccdot%20b%20%3D%20%5B%288.6%20%2B%209.3%29i%20%2B%20%285.1%20%2B%209.5%29j%5D%5Ccdot%20%289.3i%20%2B%209.5j%29%20%3D%20%2817.9i%20%2B%2014.6j%29%5Ccdot%20%289.3i%20%2B%209.5j%29%20%3D%20305.17)
d) The component of a along the direction of b is:
![a*cos(\theta) = \frac{a\cdot b}{|b|} = \frac{128.43}{\sqrt{9.3^{2} + 9.5^{2}}} = 9.66](https://tex.z-dn.net/?f=%20a%2Acos%28%5Ctheta%29%20%3D%20%5Cfrac%7Ba%5Ccdot%20b%7D%7B%7Cb%7C%7D%20%3D%20%5Cfrac%7B128.43%7D%7B%5Csqrt%7B9.3%5E%7B2%7D%20%2B%209.5%5E%7B2%7D%7D%7D%20%3D%209.66%20)
I hope it helps you!