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3 years ago

14

If fluorine was in ionic compounds with ions of each of the following atoms, how many fluorines would be needed?

1 answer:

nevsk [136]3 years ago

8 0

<span>You forgot to post the list of "following atoms", but

Fluorine ions have a -1 charge. So find the charges of the other ions and The number of Fluorines you'll need should match the </span>other ion's charge.

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Do other planets have seasons? why

Brrunno [24]

Yes because all planets when they spin arund the sun they change season ,tho they change faster or slower then ours

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3 years ago

What are subtomic particles<br>

slavikrds [6]

Answer:

In the physical sciences, subatomic particles are smaller than atoms. They can be composite particles, such as the neutron and proton; or elementary particles, which according to the standard model are not made of other particles. Particle physics and nuclear physics study these particles and how they interact.

Explanation:

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2 years ago

Read 2 more answers

The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

$R_{CM} =$ Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

$g = \frac{GM}{(d-R_{CM})^2}$

$g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}$

$g = 3.4*10^{-5}m/s^2$

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

$\omega = \frac{2\pi}{T}$

At the same time we have that centripetal acceleration is given as

$a_c = \omega^2 r$

Replacing

$a_c = (\frac{2\pi}{T})^2 r$

$a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)$

$a_c = 3.34*10^{-5}m/s^2$

3 0

3 years ago

An object starts from rest, and accelerates at 2m/s2 for 10s. How far has it gone in that time

blagie [28]

Answer:

100m

Explanation:

$s = ut + \frac{1}{2} a {t}^{2}$

u=0;t=10sec;a=2m/s²

$s = 0(10) + \frac{1}{2}(2 \times {10}^{2} )$

s=10²;100m

7 0

3 years ago

To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 1 V battery connected in series with a switch.

ki77a [65]

Answer:

0.0133A

Explanation:

Since we have two sections, for the Inductor region there would be a current $i_1$ . In the case of resistance 2, it will cross a current $i_2$

Defined this we proceed to obtain our equations,

For $i_1$ ,

$\frac{di_1}{dt}+i_1R_1 = V$

$I_1 = \frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})$

For $i_2$ ,

$I_2R_2 =V$

$I_2 = \frac{V}{R_2}$

The current in the entire battery is equivalent to,

$i_t = I_1+I_2$

$i_t = \frac{V}{R_2}+\frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})$

Our values are,

$V=1V$

$R_1 = 95\Omega$

$L= 1.5*10^{-2}H$

$R_2 =360\Omega$

Replacing in the current for t= 0.4m/s

$i=\frac{1}{360}+\frac{1}{95}(1-e^{-\frac{95*0.4}{1.5*10^{-2}}})$

$i= 0.0133A$

$i_1 = 0.01052A$

3 0

3 years ago