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lilavasa [31]
3 years ago
12

What is the name of the atomic model in which electrons are treated as waves

Physics
1 answer:
chubhunter [2.5K]3 years ago
5 0

Answer:

The atomic model in which electrons are treated as waves is called the wave mechanical model of the atom or the quantum mechanical model of the atom. Principal energy levels contains energy sublevels

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As the wavelength increases, the frequency (2 points) decreases and energy decreases. increases and energy increases. decreases
frozen [14]
Bohr's equation for the change in energy is
\Delta E= \frac{hc}{\lambda}
where
h = Planck's constant
c == the velocity of light
λ = wavelength.

The velocity is related to wavelength and frequency, f, by
c = fλ

Let us examine the given answers on the basis of the given equations.

a. As λ increases, f decreases and ΔE decreases.
     TRUE

b. As λ increases, f increases and ΔE increases.
    FALSE

c. As λ increases, f increases and ΔE decreases.
    FALSE

Answer: 
As the wavelength increases, the frequency decreases and energy decreases.

3 0
3 years ago
Read 2 more answers
a hawk flies in a horizontal arc of radius 10.3 m at a constant speed of 4.8 m/s. find its centripetal acceleration. answer in u
n200080 [17]

The hawk’s centripetal acceleration is 2.23 m/s²

The magnitude of the acceleration under new conditions is 2.316 m/s²

radius of the horizontal arc = 10.3 m

the initial constant speed = 4.8 m/s

we know that the centripetal acceleration is given by

    a_{c}  = \frac{v^{2} }{r}

   a_{c}  = 23.04/10.3

    a_{c}  = 2.23 m/s²

It continues to fly but now with some tangential acceleration

a_{t} = 0.63 m/s²

therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

so

a_{net}  =  \sqrt{a_{c} ^{2} +a_{t} ^{2}   }

a_{net}  =  \sqrt{4.97 + 0.396}

a_{net}  =  2.316 m/s²

So the magnitude of  net acceleration will become 2.316 m/s².

learn more about acceleration here :

brainly.com/question/11560829

#SPJ4

8 0
1 year ago
A space shuttle takes off from FL and circles Earth several times, finally landing in CA. While the shuttle is in flight, a phot
mixer [17]

Answer:

Both the astronauts and photographer have the same displacement

Explanation:

Displacement is the minimum distance between two point. The initial point of both the astronauts and the photographer was Florida and the final point was California. So, the minimum distance for both of the astronauts and the photographer would be the distance between Florida and California would be the same.

Hence, both the astronauts and photographer will have the same displacement.

3 0
3 years ago
A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra
bija089 [108]

Explanation:

Terminal velocity is given by:

v_t=\sqrt{\frac{2mg}{\rho C_dA}}

Here, m is the mass of the falling object, g is the gravitational acceleration,  C_d is the drag coefficient, \rho is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is 1.28\frac{kg}{m^3}.

v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}

7 0
2 years ago
the density of a steel is 7.8.find the mass of steel cube of side 10 cm. find volume of steel if the mass is 8kg​
lukranit [14]

Answer:

Mass of the steel cube =  7800 kg

Volume of the steel  = 1.025 cubic centimetre

Explanation:

Given:

The density of the steel   =  7.8

Side of the cube = 12 cm

<u>(1)The mass of steel cube :</u>

We know that,

Density  = \frac{Mass}{Volume}

We are given with density and sides of the cube

then volume of the cube

= (side)^3

= 10^3

= 1000 cubic centimetre

Now

7.8 = \frac{mass}{1000}

mass = 7.8 \times 1000

mass =  7800 kg

<u>(2)volume of steel:</u>

Given the mass  = 8 kg

Density  = \frac{Mass}{Volume}

Substituting the values

7.8 = \frac{8}{volume}

volume = \frac{8}{7.8}

volume = 1.025 cubic centimetre

3 0
2 years ago
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