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3 years ago

Like charges will exert a force of

Physics

2 answers:

natulia [17]3 years ago

7 0

Answer:

D- Repulsion

Explanation:

A positively charged object will exert a repulsive force upon a second positively charged object.

Vlad1618 [11]3 years ago

6 0

Repulsión can someone help me ? What is the summary of chapter 27 book two roads by Joseph

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What crisis is occurring in California?

nlexa [21]

Hey there!

Option A

A climate crisis is occurring in California where sudden forest fires occur due to this .

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2 years ago

Select the correct answer from each drop-down menu.

earnstyle [38]

Answer:

a transverse (sort of a plot of a sine or cosine graph, basically)

b longitudinal

c Electromagnetic (an electric wave and a magnetic wave travelling together at right angles to each other)

Explanation:

7 0

3 years ago

CHEGG You stretch a spring with spring constant k = 1.2x104 N/m to extend 6.0 cm away from its equilibrium position. How much do

lubasha [3.4K]

Answer:

The elastic potential energy of the spring change during this process is 21.6 J.

Explanation:

Given that,

Spring constant of the spring, $k=1.2\times 10^4\ N/m$

It extends 6 cm away from its equilibrium position.

We need to find the elastic potential energy of the spring change during this process. The elastic potential energy of the spring is given by the formula as follows :

$E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 1.2\times 10^4\times (0.06)^2\\\\E=21.6\ J$

So, the elastic potential energy of the spring change during this process is 21.6 J.

4 0

3 years ago

1. If I travel 50 miles in 5minutes, how fast am I going?

Anarel [89]

Answer:

10 miles a min

Explanation:

math

3 0

3 years ago

The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu

ivolga24 [154]

Answer: $0.10233nm$

Explanation:

The mean free path $\lambda$ of an atom is given by the following formula:

$\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}$ (1)

Where:

$\lambda=0.2\mu m=0.2(10)^{-6}m$

$R=8.3145J/mol.K$ is the Universal gas constant

$T=0\°C=273.115K$ is the absolute standard temperature

$d$ is the diameter of the helium atom

$N_{A}=6.0221(10)^{23}/mol$ is the Avogadro's number

$P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3}$ absolute standard pressure

Knowing this, let's find $d$ from (1), in order to find the radius $r$ of the helium atom:

$d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}$ (2)

$d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}$ (3)

$d=2.0467(10)^{-10}m$ (4)

If the radius is half the diameter:

$r=\frac{d}{2}$ (5)

Then:

$r=\frac{2.0467(10)^{-10}m}{2}$ (6)

$r=1.0233(10)^{-10}m$ (7)

However, we were asked to find this radius in nanometers. Knowing $1nm=(10)^{-9}m$ :

$r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm$ (8)

Finally:

$r=0.10233nm$ This is the radius of the helium atom in nanometers.

5 0

2 years ago