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3 years ago

Turquoise is what types of rocks sedimentary, metamorphic or igneous

Physics

1 answer:

Step2247 [10]3 years ago

3 0

Turquoise is in metamorphic and sedimentary rocks. Hope that helped

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A wire is stretched right to its breaking point by a 5000 N force. A longer wire made of the same material has the same diameter

leva [86]

Answer:

Equal to 5000N

Explanation:

The stress on the material is defined by force per unit of cross-sectional area. So it depends on the force and the diameter of the wire, which is the same for both wires. The material that defines the breaking point, is also the same. Therefore, both wires have their breaking point the same at 5000N. The wire length plays no role in here.

4 0

3 years ago

The properties of metals depend mainly on the number and arrangement of neutrons<br> True<br> False

Sergeeva-Olga [200]

Answer:

False

Explanation:

Electrons

7 0

3 years ago

What are at least three types of energy involving a microwave

timama [110]

I'm pretty sure what you are trying to ask for is radiative energy, light energy, and electronic energy.
Radiative since the microwave is releasing radiation,
Light since there is light inside the microwave,
Electronic since it is plugged in and uses electricity.
You can also use sound, but I don't think every microwave makes sound.

8 0

3 years ago

A rock with a mass of 8 kg falls straight down from a height of 7 m. What work is done?

fenix001 [56]

F=MA
F=(8 kg)(9.8 m/s)
F= 78.4 N
W=FD
W=(78.4 N)(7 m)
W=548.8 J
How this helps

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