3 years ago

An electron in a mercury atom drops from energy level f to energy level c by emitting a photon having an energy of

Physics

2 answers:

arlik [135]3 years ago

6 0

Answer: it’s 2

Explanation: you can find the answer on the physics reference sheet

Julli [10]3 years ago

5 0

Answer:

The answer is 2.84 eV

Explanation:

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Fluid pressure ____________ as elevation ____________.

ad-work [718]

Y = 2x $y =2x \sqrt{x}$

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3 years ago

A 40 N crate rests on a rough horizontal floor. A 12 N horizontal force is then applied to it. If the coecients of friction are

Crank

Answer:

Frictional force, f = 20 N

Explanation:

It is given that,

Weight of the crate, W = 40 N

Horizontal force acting on the crate, F = 12 N

The coefficient of static friction, $\mu_s=0.5$

The coefficient of kinetic friction, $\mu_s=0.4$

Let f is the frictional force acting on the crate. Friction is an opposing force. It is equal to the product of coefficient friction and the normal force. It is given by :

$f=\mu_s\times N$

$f=0.5\times 40$

f = 20 N

So, the frictional force on the crate is 20 N. Hence, this is the required solution.

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4 years ago

an airplane travels down a runway for 4.0 seconds with an acceleration of 9.0m/s. what is its change in velocity during this tim

finlep [7]

A=9m/s^2, u=0, t=4 secs
by the equations of motion
a = $\frac{v-u}{t}$
9 = $\frac{v}{4}$
∴v = 36m/s

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3 years ago

How is electron movement related to the bonding in sodium chloride?

irina1246 [14]

I think its c sorry if I’m wrong

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3 years ago

Read 2 more answers

A piece of amber is charged by rubbing with a piece of fur. If the net excess charge on the fur is 8.9 nC ( 8.9 10-9 C), how man

Alika [10]

Answer:

Number of electrons, $n=5.56\times 10^{10}$

Explanation:

Given that,

Charge on the fur, $q=8.9\ nC=8.9\times 10^{-9}\ C$

A piece of amber is charged by rubbing with a piece of fur. We need to find the electrons were added to the amber. It can be calculated using the quantization of charge as

q = n × e

Where

n is the number of electrons

e is the charge on electron

$n=\dfrac{q}{e}$

$n=\dfrac{8.9\times 10^{-9}}{1.6\times 10^{-19}}$

$n=5.56\times 10^{10}\ electrons$

So, $n=5.56\times 10^{10}\ electrons$ number of electrons are added to the amber. Therefore, this is the required solution.

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4 years ago