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SOVA2 [1]
3 years ago
9

Why is it so important to avoid pollution of freshwater ecosystems? A. Freshwater ecosystems contain more species than marine ec

osystems contain. B. Freshwater ecosystems are the source of most of our drinking water. C. There are more freshwater ecosystems than there are marine ecosystems. D. Freshwater organisms are more important than marine organisms are.
Physics
2 answers:
Mrrafil [7]3 years ago
5 0

The answer is B because it literally says it right there, and I don't wanna make it impossibly confusing for a middle school student, lol.

makkiz [27]3 years ago
3 0

it is b!!!!!!!!!!!!!!!!!!!!!!

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Because AC emf is a sine wave
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2 years ago
If a bust starts to move and its velocity becomes 90 km after 8 seconds . calculate its acceleration answer it quick please
kherson [118]

Answer:

a = 3.125 [m/s^2]

Explanation:

In order to solve this problem, we must use the following equation of kinematics. But first, we have to convert the speed of 90 [km/h] to meters per second.

90\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}   \\= 25 \frac{m}{s}

v_{f} =v_{i} + (a*t)

where:

Vf = final velocity = 25 [m/s]

Vi = initial velocity = 0

a = acceleration [m/s^2]

t  = time = 8 [s]

The initial speed is zero as the bus starts to koverse from rest. The positive sign of the equation means that the bus increases its speed.

25 = 0 + a*8

a = 3.125 [m/s^2]

3 0
3 years ago
What is the abbreviation for mojave desert?
vagabundo [1.1K]
There is no abbreviation
3 0
3 years ago
A long, East-West-oriented power cable carrying an
Alla [95]

Answer:

200A

Explanation:

Given that

the distance between earth surface and power cable d = 8m

when the current is flowing through cable , the magnitude flux density at the surface is 15μT

when the current flow throught is zero the magnitude flux density at the surface is 20μT

The change in flux density due to the current flowing in the power cable is

B = 20μT - 15μT

B =5μT -----(1)

The expression of magnitude flux density produced by the current carrying cable is

B=\frac{\mu_0I}{2\pi d}-----(2)

Substitute the value of flux density

B from eqn 1 and eqn 2

\frac{\mu_0I}{2\pi d}=5\times 10^-^6\\\\\frac{(4\pi \times 10^-^7)I}{2 \pi (8)} =5\times 10^-^6\\\\I=200A

Therefore, the magnitude of current I is 200A

8 0
2 years ago
Can someone explain to me #4.
nalin [4]
A) We balance the masses: 4(1.00728) vs 4.0015 + 2(0.00055)4.02912 vs. 4.0026This shows a "reduced mass" of 4.02912 - 4.0026 = 0.02652 amu. This is also equivalent to 0.02652/6.02E23 = 4.41E-26 g = 4.41E-29 kg.
b) Using E = mc^2, where c is the speed of light, multiplying 4.41E-29 kg by (3E8 m/s)^2 gives 3.96E-12 J of energy.
c) Since in the original equation, there is only 1 helium atom, we multiply the energy result in b) by 9.21E19 to get 3.65E8 J of energy, or 365 MJ of energy.
4 0
3 years ago
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