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Feliz [49]
3 years ago
10

Upwelling _____.

Physics
2 answers:
Assoli18 [71]3 years ago
8 0
- - A up welling is a solution appearing and moving towards the ocean surface meaning as, its going up,
Therefore, C <span>happens when water rises to replace moving surface water</span>
ANEK [815]3 years ago
8 0

The answer is C, happens when water rises to replace moving surface water.

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What is keeping all variables the same except for the one being tested an example of?
stepan [7]

Answer:

A fair test.

Explanation:

Hi, a fair test is used to do scientifically valuable experiments, is a controlled investigation to answer a scientific question.

In a fair test two or more things are compared.

It consists in changing only one factor (the one bieng tested) and keeping all the other conditions the same during an experiment.

The factor is called a variable.

8 0
3 years ago
Read 2 more answers
A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 6.0 oz. Aluminum has a density of 2.70 g/cm3. Wh
SIZIF [17.4K]

Answer:

1.36 x 10^-3 cm

Explanation:

Area = 50 ft^2 = 46451.5 cm^2

mass = 6 oz = 170.097 g

density = 2.70 g/cm^3

Let t be the thickness of foil in cm.

mass = volume x density

mass = area x thickness x density

170.097 = 46451.5 x t x 2.70

t = 1.36 x 10^-3 cm

Thus, the thickness of aluminium foil is 1.36 x 10^-3 cm.

3 0
3 years ago
You are operating a pwc in an area where swimmers are in the water. When must you slow your pwc to "slow, no wake speed"?.
Lostsunrise [7]

You should slow your pwc to "slow, no wake speed" when within 100 feet of anchored vessels or non-motorized craft.

<h3>What is Slow-no-wake?</h3>

This is the process of operating a personal watercraft at the slowest possible speed.

This helps to maintain steerage which prevents different forms of accident or risks when in motion in the water.

Read more about Slow-no-wake here brainly.com/question/10410716

#SPJ1

4 0
2 years ago
Help<br> How much current will flow when a 120 V power supply is connected to a 30<br> resistor ?
AVprozaik [17]
Current= voltage divided by resistance
120/30=4
7 0
3 years ago
A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri
GREYUIT [131]

Answer:

|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}

Now, substitute this into 'dF':

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)

Now we can integrate dF over the rod.

\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)

4 0
3 years ago
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