<span>Crust. The thin solid outermost layer of Earth. ...Asthenosphere. The lower layer of the crust. ...Lithosphere.Plasticity: is solid but still being able to. flow without being a liquid.The cool, rigid outermost layer of the Earth. ...<span>the solid part of the earth consisting of the crust and outer mantle.</span></span>
The answer to this statement is true!
Answer:
Explanation:
a ) Slit separation d = .1 x 10⁻³ m
Screen distance D = 4 m
wave length of light λ = 650 x 10⁻⁹ m
Width of central fringe = λ D / d
= 
= 26 mm
b ) Distance between 1 st and 2 nd bright fringe will be equal to width of dark fringe which will also be equal to 26 mm
c ) Angular separation between the central maximum and 1 st order maximum will be equal to angular width of fringe which is equal to
λ / d
= 
= 6.5 x 10⁻³ radian.
Answer:
The COP of the system is = 4.6
Explanation:
Given data
Higher pressure = 1.8 M pa
Lower pressure = 0.12 M pa
Now we have to find out high & ow temperatures at these pressure limits.
Higher temperature corresponding to pressure 1.8 M pa
°c = 335.9 K
Lower temperature corresponding to pressure 0.2 M pa
°c = 262.9 K
COP of the system is given by
COP = 4.6
Therefore the COP of the system is = 4.6
The complete ionization of KBr into its constituents
is:<span>
<span>KBr (s) --->
K+ (aq) + Br- (aq)</span></span>
<span>
During electrolysis, oxidation takes place at the anode electrode. This means
that an ion is stripped off its electron hence becoming more positive:
<span>2 Br- (aq) --->
Br2 (g) + 2e- </span></span>
We can see that Bromine gas Br2 is evolved at the anode.
<span>
<span>Meanwhile at the cathode, the reduction reaction occurs.
Which means that the electron from the anode electrode is used to make an ion
more negative:
<span>2K+ (aq) + 2e- ---> 2K (s) </span></span>
Hence, through reduction, solid potassium is deposited on the
plate.</span>
Half reactions:
<span>Anode: 2 Br- (aq) --->
Br2 (g) + 2e- </span>
<span>Cathode: 2K+ (aq) + 2e-
---> 2K (s) </span>
