Answer:
wavelength decreases and frequency increase
Explanation:
the higher the wavelength the smaller the frequency , the smaller the wavelength the higher the frequency
Thank you for posting your question here at brainly. Below is the solution. I hope the answer will help.
<span>Cl^- 1s^2 2s^2p^6 3s^2 3p^6 1s^2 2s^2p^6 S = 10; 3s^2 3p^6 S = 0 </span>
<span>Zeff = Z-S = 17- 10 =7 </span>
<span>K^+ 1s^2 2s^2p^6 3s^2 3p^6; 1s^2 2s^2p^6 S = 10; 3s^2 3p^6 S = 0 </span>
<span>Zeff = Z-S = 19- 10 = 9
</span>
S = 2 + 6.8 + 2.45 = 11.25
<span>Zeff(Cl^-) = 17 – 11.25 = 5.75 </span>
<span>K^+ 1s^2 2s^2p^6 3s^2 3p^6 same S as for Cl^- but Z increases by 2 hence </span>
<span>Zeff(K^+) = 19 - 11.25 = 7.75</span>
A) 
Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:
Parallel:
(1)
where
m is the mass
g = 9.8 m/s^2 the acceleration of gravity

is the coefficient of friction
R is the normal reaction
a is the acceleration
Perpendicular:
(2)
From (2) we find

And substituting into (1)

Solving for a,

B) 5.94 m/s
We can solve this part by using the suvat equation

where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
Here we have
v = ?
u = 0 (it starts from rest)

s = 8.70 m
Solving for v,

Answer:
lives, forces and motion make things move and stay still. Pushing and pulling are examples of forces that can sped things up or slow things down. There are two types of forces, at a distance force and contact forces..
Answer:
elastic partial width is 2.49 eV
Explanation:
given data
ER E = 250 eV
spin J = 0
cross-section magnitude σ = 1300 barns
peak P = 20ev
to find out
elastic partial width W
solution
we know here that
σ = λ²× W / ( E × π × P ) ...................1
put here all value
σ = (0.286)² × W ×
/ ( 250 × π × 20 )
1300 ×
= (0.286)² × W ×
/ ( 250 × π × 20 )
solve it and we get W
W = 249.56 ×
so elastic partial width is 2.49 eV