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3 years ago

3) If an object has a net negative charge of 4.0 Coulombs, the object possesses

Physics

1 answer:

Pani-rosa [81]3 years ago

3 0

The answer to this question is option 2

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Light hits a mirror at a 45Â° angle. It will be reflected at an angle _____.

Eddi Din [679]

It will be equal to 45Â°

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3 years ago

How can someone, like myself, share roughly 50% of their DNA with a 1st Cousin?

iragen [17]

Answer:

that's your 1st cousin which is your mom or dad's sibling's child sooo when you're that close and that related i would say that that would be the reason why.

Explanation:

hoped that helped!!

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3 years ago

As the train in the image moves to the right how does the train horn sound to person a?

alina1380 [7]

Answer:

Explanation:

Person A's velocity relative to the train is 0. Therefore, the pitch of the horn will not change.

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3 years ago

Read 2 more answers

The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 542 nm. What is t

Veseljchak [2.6K]

Answer:

2.295 eV

Explanation:

maximum wavelength, λ = 542 nm = 542 x 10^-9 m

The work function of the metal is defined as the minimum amount of energy falling on the metal so that the photo electrons just ejects the surface of metal.

$W_{o}=\frac{hc}{\lambda }$

where, h is the Plank's constant and c be the speed of light

h = 6.634 x 10^-34 Js

c = 3 x 10^8 m/s

$W_{o}=\frac{6.634\times 10^{-34}\times 3\times 10^{8}}{542\times10^{-9} }$

$W_{o}=3.67\times 10^{-19} J=\frac{3.67\times 10^{-19}}{1.6\times 10^{-19}} eV$

Wo = 2.295 eV

Thus, the work function of this metal is 2.295 eV.

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3 years ago

The atomic radii of Mg2+ and F- ions are 0.072 and 0.133 nm, respectively.(a) Calculate the force of attraction between these tw

timurjin [86]

Answer:

$1.09527\times 10^{-8}\ N$

Explanation:

$q_1$ = Mg ion = $+2q$

$q_2$ = F ion = $-q$

q = Charge of electron = $1.6\times 10^{-19}\ C$

r = Distance between ions = $0.072+0.133\ nm$

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electrical force is given by

$F=-\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=-\dfrac{8.99\times 10^9\times 2\times 1.6\times 10^{-19}\times -1\times 1.6\times 10^{-19}}{[(0.072+0.133)\times 10^{-9}]^2}\\\Rightarrow F=1.09527\times 10^{-8}\ N$

The attractive force is $1.09527\times 10^{-8}\ N$

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4 years ago