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sattari [20]
2 years ago
7

Erwin Schrödinger developed the quantum model of the atom. What scientific knowledge was necessary for Schrödinger’s work?

Physics
1 answer:
Stella [2.4K]2 years ago
7 0

Answer:

Erwin Schrödinger developed the quantum model of the atom. What scientific knowledge was necessary for Schrödinger's work? The discovery of the electron.

Explanation:

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A sled is accelerating down a hill at a rate of 1 m s2 . If the mass of the sled is suddenly cut in half and the net force on th
mihalych1998 [28]
We have that F=ma from the 2nd Newton law where F is the force, m is the mass and a is the acceleration. Suppose we have that F' is the new force and m' is the new mass. Then, we have that a'=F'/m' still, by rearranging Newton's law. We are given that F'=2F and m'=m/2. Hence,
a'= \frac{2F}{ \frac{m}{2} } = \frac{4F}{m} = 4\frac{F}{m}
But now, we have from F=ma, that a=F/m and we are given that a=1m/s^2.
We can substitute thus, a'=4a=4*1m/s^2=4m/s^2.
4 0
3 years ago
Why do we not feel air pressure?
professor190 [17]

Answer:

Explanation:

Is beacuse of the air within our bodys is exerting the same pressure out wards so tjere is no pressure difference

8 0
3 years ago
Read 2 more answers
A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an
maksim [4K]

Answer:

The  value is   \alpha =  0.002 Np/m

Explanation:

From the question we are told that

  The first amplitude of the wave is  E_{max}1 =  98.02 \  V/m

  The first  depth  is  D_1 =  10 \  m

   The second amplitude is  E_{max}2 =  81.87 \  (V/m)

   The second depth is D_2 = 100 \ m

Generally from the spatial wave equation we have

   v(x) =  Ae^{-\alpha d}cos(\beta x  + \phi_o)

=>       \frac{v(x)}{v(x)} =\frac{  Ae^{-\alpha d}cos(\beta x  + \phi_o)}{ Ae^{-\alpha d}cos(\beta x  + \phi_o)}

So considering the ratio of the equation for the  two depth

\frac{A}{A_S}  =  \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}

=>   \frac{98.02}{81.87}  =  \frac{e^{-10 \alpha }}{e^{-100 \alpha }}

=>   \alpha  =  \frac{0.18}{90}

=>    \alpha =  0.002 Np/m

       

4 0
3 years ago
What magnification would be obtained if an eyepiece with a focal length of 0.38 m was placed on telescope?
weqwewe [10]

Answer:

This question is incomplete

Explanation:

This question is incomplete because the telescope's focal length was not provided. The formula to be used here is

Magnification = telescope's focal length/eyepiece's focal length

The eyepiece's focal length was provided in the question as 0.38 m.

NOTE: Magnification can be described as the power of an instrument (in this case telescope) to enlarge an object. It has no unit and thus the two focal lengths mentioned in the formula above must be in the same unit (preferably meters since one of them is in meters already).

7 0
2 years ago
How long( in hours, will it take for 500 000 C of charge to flow through a diode if it requires
Alina [70]

Answer:

277.78 hours

Explanation:

The formula for calculating the amount of charge is expressed as;

Q = It

I is the current

t is the time

Given

I =0.05A

Q = 50,000C

Required

Time t

Recall that: Q = It

t = Q/I

t = 50,000/0.05

t = 1,000,000secs

Convert to hours

1,000,000secs  = 1,000,000/3600

1,000,000secs = 277.78 hours

Hence it will take 277.78 hours for the charge to flow through the diode

6 0
3 years ago
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