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DIA [1.3K]
3 years ago
5

What two quantities affect the force of attraction between two charges

Physics
2 answers:
miss Akunina [59]3 years ago
5 0
1. the amount of electric charge that they carried
2. the distance between them
 
Orlov [11]3 years ago
5 0
1).  the product of their charges

2).  the distance between their centers of charge
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Red Riding Hood sat down to rest and placed her 1.20-kg basket of goodies beside her. A wolf came along, spotted the basket, and
anygoal [31]

Answer:

  117.6°

Explanation:

The vertical component of a force directed at some angle α from the vertical is ...

  F·cos(α)

We want the vertical components of the wolf's force (Fw) and Red's force (Fr) to total zero. So for some angle from vertical α, Red's force will satisfy ...

  Fw·cos(25°) + Fr·cos(α) = 0

  cos(α) = -Fw/Fr·cos(25°) ≈ -(6.4 N)/(12.5 N)·0.906308 ≈ -0.464030

  α ≈ arccos(-0.464030) ≈ 117.6°

Red was pulling at an angle of about 117.6° from the vertical.

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<em>Additional comment</em>

That's about 27.6° below the horizontal.

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3 years ago
What happens when the dew point and the temperature are the same?
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The value of relative humidity becomes 100% leading to condensation of water vapor in the air into water droplets or water (dew)
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3 years ago
A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are
slava [35]

Answer:

The magnitude and direction of the acceleration of the particle is a= 0.3296\ \hat{k}\ m/s^2

Explanation:

Given that,

Mass m = 1.81\times10^{-3}\ kg

Velocity v = (3.00\times10^{4}\ m/s)j

Charge q = 1.22\times10^{-8}\ C

Magnetic field B= (1.63\hat{i}+0.980\hat{j})\ T

We need to calculate the acceleration of the particle

Formula of the acceleration is defined as

F = ma=q(v\times B)

a =\dfrac{q(v\times B)}{m}

We need to calculate the value of v\times B

v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})

v\times B=4.89\times10^{4}

Now, put the all values into the acceleration 's formula

a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}

a= -0.3296\ \hat{k}\ m/s^2

Negative sign shows the opposite direction.

Hence, The magnitude and direction of the acceleration of the particle is a= 0.3296\ \hat{k}\ m/s^2

7 0
3 years ago
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What is atomic theory and how has it changed over time?
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6 0
3 years ago
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Calculate the de Broglie wavelength of (a) a mass of 1.0 g traveling at 1.0 m s−1 , (b) the same, traveling at 1.00 × 105 km s−1
lesantik [10]

Answer:

a)\lambda=6.63\times10^{-31}m

b)\lambda=6.63\times10^{-39}m

c)\lambda=9.97\times10^{-11}m

d)\lambda=4.03\times10^{-36}m

e)λ=∞

Explanation:

De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

\lambda=\frac{h}{P}=\frac{h}{mv}

with h the Plank's constant (6.63\times10^{-34}\frac{m^{2}kg}{s}) and P the momentum of the object that is mass (m) times velocity (v).

a)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}

\lambda=6.63\times10^{-31}m

b)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}

\lambda=6.63\times10^{-39}m

c)\lambda=\frac{6.63\times10^{-34}}{(6.65\times10^{-27}*1000)}

\lambda=9.97\times10^{-11}m

d)\lambda=\frac{6.63\times10^{-34}}{(74*2.22)}

\lambda=4.03\times10^{-36}m

e) \lambda=\frac{6.63\times10^{-34}}{(74*0)}

λ=∞

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3 years ago
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