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nata0808 [166]
3 years ago
10

Which group has the Same electron shells? A Titanium and Gold B Selenium and Tin C Calcium and Magnesium D Lithium and Boron

Physics
1 answer:
Anna11 [10]3 years ago
6 0

im not 100 percent sure but i think its b


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How can citizen -scientists help with the prediction of future earthquakes
iVinArrow [24]

Answer:

Avoid downed power lines and stay away from buildings and bridges from which heavy objects might fall during an aftershock. Stay away until local officials tell you it is safe. A tsunami is a series of waves that may continue for hours. Do not assume that after one wave the danger is over.

3 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
On a cool morning, Uyen’s breath can form a cloud when she breathes out. Which changes of state are most responsible for Uyen se
zhuklara [117]
It's cold outside, the water vaper in your breath condenses into tiny droplets of liquid water and ice that you can see.
3 0
3 years ago
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An object starts at 4m/s and accelerates to 6m/s in 10 seconds. Find its displacement.
Natalka [10]

Answer:

Explanation:

The 2 equations we need here are, first:

a=\frac{v_f-v_0}{t} and then once we solve for the acceleration here:

v^2=v_0^2+2aΔx

Solving for acceleration:

a=\frac{6-4}{10}=\frac{1}{5}\frac{m}{s^2} and now we will use that in the other equation:

6^2=4^2+2(\frac{1}{5})Δx and

36 = 16 + \frac{2}{5}Δx and

20 = \frac{2}{5}Δx and

\frac{5}{2}(20)= Δx so

Δx = 50 m

6 0
3 years ago
A 1.0-m long wire is carrying a certain amount of current. The wire is placed perpendicular to a magnetic field of strength 0.20
Marysya12 [62]

For a current-carrying wire running perpendicular to a magnetic field, the magnetic force acting on the wire is given by:

F = ILB

F = magnetic force, I = current, L = wire length, B = magnetic field strength

Given values:

F = 0.60N, L = 1.0m, B = 0.20T

Plug in and solve for I:

0.60 = I(1.0)(0.20)

I = 3.0A

5 0
3 years ago
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