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3 years ago

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What is the primary body that influences Earth's ocean tides? A. the Moon B. Venus C. Mars D. the Sun

2 answers:

Angelina_Jolie [31]3 years ago

7 0

The asnwer is A. The moon

madreJ [45]3 years ago

6 0

A~ the moon

why?
i saw it in a movie :D

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A student is playing with a pendulum. He gives the ball a push and watches the ball as it swings. After a while, the ball stops

julia-pushkina [17]

A because it’s not good

4 0

3 years ago

The mass of a sulfur atom is 32.0 u, and the mass of a fluorine atom is 19.0 u. What is the mass of a mole of sulfur hexafluorid

givi [52]

<u>Answer:</u> The mass of sulfur hexafluoride is 146 u.

<u>Explanation:</u>

Mass of a mole of a substance is defined as the molar mass of that substance.

It is defined as the sum of every component each multiplied by the times they appear in that substance.

We are given:

Mass of sulfur atom = 32.0 u

Mass of fluorine atom = 19.0 u

The chemical formula for sulfur hexafluoride is $SF_6$

So, the molar mass of $SF_6=[32+(6\times 19)=146u$

Hence, the mass of sulfur hexafluoride is 146 u.

5 0

3 years ago

Consider a uniformly charged thin-walled right circularcylindrical shell having total charge Q, radiusR, and height h. Determine

Vinil7 [7]

Answer:

$\frac{k_eQ}{2h}$

Explanation:

The question is missing an image. I have added this as an attachment to my answer.

Given;

Q = total charge

R = radius of cylindrical shell

h = height of cylindrical shell

d = distance of point from the right side of the cylinder

Let the thickness of the cylindrical shell be $dx$ , and the charge $\frac{Qdx}{h}$ ,

Now, using the formula for finding the electric field due to a ring at a chosen point:

$dE = \frac{k_ex}{(x^2 + R^2)^{\frac{3}{2}}} {\frac{Qdx}{h}i}$

where $x$ = center of the ring to the point

$k_e$ = electrostatic constant

We integrate on both sides from the limits $d$ to $d + h$ in order to determine the electric field at the point $E$

$\int\limits dE$ = $\int\limits^{d + h}_d {\frac{k_eQxdx}{h(x^2 + R^2)^{\frac{3}{2}}}i}$

$E = \int\limits^{d + h}_d {\frac{k_eQdx}{h(x^2 + R^2)^{\frac{3}{2}}}i} = \frac{k_eQ}{2h}$

6 0

3 years ago

Simple Question! Help!

Korvikt [17]

<em>60km</em><em>/</em><em>hr</em><em>.</em><em>.</em>

<em>I</em><em> </em><em>think</em><em> </em><em>so</em><em>.</em><em>.</em><em>.</em><em>.</em>

8 0

3 years ago

Read 2 more answers

A nonconducting sphere has radius R = 2.81 cm and uniformly distributed charge q = +2.35 fC. Take the electric potential at the

Sladkaya [172]

Answer:

(a). The electric potential at 1.650 cm is $-1.219\times10^{-4}\ V$ .

(b). The electric potential at 2.81 cm is $-3.759\times10^{-4}\ V$ .

Explanation:

Given that,

Radius of sphere R=2.81 cm

Charge = +2.35 fC

Potential at center of sphere

$V = 0$

(a). We need to calculate the potential at a distance r = 1.60 cm

Using formula of potential difference

$V_(r)-V_(0)=-\int_{0}^{r}{E(r)}dr$

$V_{r}-0=-\int_{0}^{r}{\dfrac{qr}{4\pi\epsilon_{0}R^3}}dr$

$V_{r}=-(\dfrac{qr^2}{8\pi\epsilon_{0}R^3})_{0}^{1.60\times10^{-2}}$

$V_{r}=-(\dfrac{2.35\times10^{-15}\times(1.60\times10^{-2})^2}{8\times\pi\times8.85\times10^{-12}\times(2.81\times10^{-2})^3})$

$V_{r}=-0.00012190\ V$

$V_{r}=-1.219\times10^{-4}\ V$

The electric potential at 1.650 cm is $-1.219\times10^{-4}\ V$ .

(b). We need to calculate the potential at a distance r = R

Using formula of potential difference

$V_{R}=-\dfrac{2.35\times10^{-15}}{8\pi\times8.85\times10^{-12}\times2.81\times10^{-2}}$

$V_{R}=-0.0003759\ V$

$V_{R}=-3.759\times10^{-4}\ V$

The electric potential at 2.81 cm is $-3.759\times10^{-4}\ V$ .

Hence, This is the required solution.

5 0

3 years ago