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pashok25 [27]
3 years ago
10

Which of the following would a geologist rely on to get data about nutrients cycle in the water

Physics
2 answers:
Novosadov [1.4K]3 years ago
8 0
The answer would be : <span>hydrologists.

Hope this helps !

Photon</span>
mrs_skeptik [129]3 years ago
8 0
There's nothing on your list of choices that he could consult for that information.
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Using Gauss's law, calculate the electric field at a point distance s from a long wire bearing uniform charge density. i need he
11111nata11111 [884]

Answer:

E = 2k  \frac{\lambda}{ r}

Explanation:

Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.

We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.

            Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀

tells us that the linear charge density is

            λ = q_ {int} /l

            q_ {int} = l λ

we substitute

            E A = l λ /ε₀

is area of ​​cylinder is

           A = 2π r l

we substitute

            E = \frac{ l \ \lambda}{ \epsilon_o \ 2\pi  \ r \ l }

             E = \frac{\lambda}{ 2\pi  \epsilon_o \ r}

the amount

            k = 1 / 4πε₀

            E = 2k  \frac{\lambda}{ r}

5 0
3 years ago
A 0.2 kg block sliding on a horizontal table slows down from 25 m/s to 20 m/s. How much energy does the block lose due to fricti
Papessa [141]

Answer:

the kinetic energy lost due to friction is 22.5 J

Explanation:

Given;

mass of the block, m = 0.2 kg

initial velocity of the block, u = 25 m/s

final velocity of the block, v = 20 m/s

The kinetic energy lost due to friction is calculated as;

\Delta K.E= K.E_f - K.E_i\\\\\Delta K.E= \frac{1}{2}mv^2 -  \frac{1}{2}mu^2\\\\\Delta K.E= \frac{1}{2}m(v^2 -u^2)\\\\\Delta K.E= \frac{1}{2} \times 0.2 (20^2 - 25^2)\\\\\Delta K.E= -22.5 \ J

Therefore, the kinetic energy lost due to friction is 22.5 J

7 0
2 years ago
Read 2 more answers
The law of universal gravitation states that a gravitational force exists between all objects in the universe. This force is dir
r-ruslan [8.4K]

Answer:

D. It has been demonstrated to be without exception under certain stated conditions.

Explanation:

A principle is simply a proposition based on some results from some experiments. A principle becomes a law when it gains strength. That is when other scientists support and back it.

A scientific law is a statement that describes a natural phenomenon and is not contradicted by repeated experiments over the time.

Sir Isaac Newton's law of gravitation stated "A gravitational force exists between all objects in the universe. This force is directly proportional to the masses of the two objects and inversely proportional to the distance between them."

His law explained that every body in the universe attracts every other body. It explained that it is not just Earth that pulls us towards it. But even sun and moon has their own gravitational pull and so does all other objects. It is just that the Gravitational pull of some objects is negligible.

Newton's law of Gravitation is considered a universal law because it is applicable to almost all the bodies in the universe and is demonstrated to be correct for them under certain conditions.

5 0
3 years ago
Read 2 more answers
A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co
algol [13]

Answer:

\displaystyle X_T=66.6\ km

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

\displaystyle V_f=V_o+a\ t

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of 0.987 m/s^2 for 182 seconds. Then it brakes with a constant acceleration of -0.321 m/s^2 until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

\displaystyle X=V_o\ t+\frac{a\ t^2}{2}

Our data is

\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec

Let's compute the first distance X1

\displaystyle X_1=0+\frac{0.987\times 182^2}{2}

\displaystyle X_1=16,346.7\ m

Now, we find the speed at the end of the first period of time

\displaystyle V_{f1}=0+0.987\times 182

\displaystyle V_{f1}=179.6\ m/s

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0

\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}

\displaystyle t=559.5\ sec

Computing the second distance

\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}

\displaystyle X_2=50,243.2\ m

The total distance is

\displaystyle X_t=x_1+x_2=16,346.7+50,243.2

\displaystyle X_t=66,589.9\ m

\displaystyle \boxed{X_T=66.6\ km}

8 0
3 years ago
How are thunderstorms kept alive?
Burka [1]

Answer:

E step leader

Explanation:

5 0
2 years ago
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