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3 years ago

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Using your Periodic Table, which element below has the smallest atomic radius? A.) Sodium, B.) Chlorine, C.) Phosphorus, D.) Iro

n

2 answers:

egoroff_w [7]3 years ago

6 0

Explanation:

Chlorine is the smallest atomic radius

Kaylis [27]3 years ago

4 0

Chlorine is the smallest because up right extreme of the periodic table

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What are the three properties of a magnet?

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2. Magnets attract or repel other magnets.

3. In addition one part of a magnet will always point north when allowed to swing freely.

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sa është syprina e gjurmës të këmbës të njeriut nëse njeriu ri në këmbë pa lëvizur dhe ka masë prej 84kg shtypja është 16.8pa

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3 years ago

A small glass bead has been charged to 20 nC. What is the magnitude of acceleration in m/s^2 of an electron that is 1.0 cm from

MAVERICK [17]

Answer:

The acceleration is 3.16x10¹⁷ m/s².

Explanation:

First, we need to find the magnitude of the Coulombs force (F):

$|F| = \frac{Kq_{1}q_{2}}{d^{2}}$

<u>Where</u>:

K is the Coulomb constant = 9x10⁹ Nm²/C²

q₁ is the charge = 20x10⁻⁹ C

q₂ is the electron's charge = -1.6x10⁻¹⁹ C

d is the distance = 1.0 cm = 1.0x10⁻² m

$|F| = \frac{Kq_{1}q_{2}}{d^{2}} = \frac{9\cdot 10^{9}Nm^{2}/C^{2}*20 \cdot 10^{-9} C*(-1.6\cdot 10^{-19} C)}{(0.01 m)^{2}} = 2.88 \cdot 10^{-13} N$

Now, we can find the acceleration:

$a = \frac{F}{m} = \frac{2.88 \cdot 10^{-13} N}{9.1 \cdot 10^{-31} kg} = 3.16 \cdot 10^{17} m/s^{2}$

Therefore, the acceleration is 3.16x10¹⁷ m/s².

I hope it helps you!

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3 years ago

How would you classify hydrogen

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4 years ago

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An armada of spaceships that is 1.60 ly long (in its rest frame) moves with speed 0.800c relative to the ground station in frame

miskamm [114]

Answer:

(a). The trip take a time is 3.61 year.

(b). The trip take a time is 3.96 year.

(c). The trip take a time is 8.73 year.

Explanation:

Given that,

Length = 1.60 ly

Speed of spaceships= 0.800c

Speed of messenger = 0.910 c

(a). We need to calculate the velocity of armada

Using formula of velocity

$v'=\dfrac{v-v_{m}}{1-\dfrac{vv_{m}}{c^2}}$

Put the value into the formula

$v'=\dfrac{0.800c-0.910c}{1-\dfrac{0.800\times0.910 c^2}{c^2}}$

$v'=\dfrac{-0.11c}{1-0.800\times0.910}$

$v'=-0.404c$

We need to calculate the length

Using formula of length

$l=l'\sqrt{1-\dfrac{v'^2}{c^2}$

Put the value into the formula

$l=1.60\sqrt{1-(-0.404)^2}$

$l=1.46\ ly$

We need to calculate the length of the trip

Using formula of time

$t=\dfrac{l}{|v'|}$

Put the value into the formula

$t=\dfrac{1.46}{0.404}$

$t=3.61\ year$

(b). If the armada's rest frame

We need to calculate the length

Using formula of length

$l=l'\sqrt{1-\dfrac{v'^2}{c^2}$

Put the value into the formula

$l=1.60\sqrt{1-(0)^2}$

$l=1.60\ ly$

Using formula of time

$t=\dfrac{l}{|v'|}$

Put the value into the formula

$t=\dfrac{1.60}{0.404}$

$t=3.96\ year$

(c). If an observer in frame S

We need to calculate the length

Using formula of length

$l=l'\sqrt{1-\dfrac{v'^2}{c^2}$

Put the value into the formula

$l=1.60\sqrt{1-(0.800)^2}$

$l=0.96\ ly$

We need to calculate the time

$t=\dfrac{0.96}{0.910-0.800}$

$t=8.73\ year$

Hence, (a). The trip take a time is 3.61 year.

(b). The trip take a time is 3.96 year.

(c). The trip take a time is 8.73 year.

8 0

3 years ago