What practice by retreating Soviet troops prevented resource gains for German troops? The Soviets froze the oil inside the military tanks. The Soviets initiated a blitzkrieg against the Germans. The Soviets enacted a scorched earth policy. The Soviets cut off the supply lines from the Axis powers.

Answer:

The grooves behind the chain have been cut over decades by motion of the chain as hikers grabbed it, and the potholes were cut by water swirling rocks around during the rare floods over much longer times.

Explanation:

In the question above, we can see that the physical action of some factors found in this environment, such as the movement of the currents and the movement of stones in the water, which managed to cause some changes in the environment, building a totally new scenario, which would not occur if these factors did not exist.

By reading the question and imagining the scenario described, we can conclude that the grooves behind the chain have been cut over decades by motion of the chain as hikers grabbed it, and the potholes were cut by water swirling rocks around during the rare floods over much longer times.

Answer:

The explanation of this questions is given below in the explanation section

Explanation:

 In this question, different statements are given against their corresponding items randomly. The solution to this question requires that match statement against given item.

In the attached picture, a table is drawn, which has 2 columns. statement and matching item. In statement column, statement are given while the items column contains the matching item against the given statement.          

Answer: The ratio of \frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}  in the cell is 4.07\times 10^{-4}

Explanation:

The half reactions for the cell is:  

Oxidation half reaction (anode):  Sn(s)\rightarrow Sn^{2+}+2e^-

Reduction half reaction (cathode):  Sn^{2+}+2e^-\rightarrow Sn(s)

In this case, the cathode and anode both are same. So, E^o_{cell} will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = 0.10 V

\frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}  = ?

Putting values in above equation, we get:

0.10=0-\frac{0.0592}{2}\log \frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}

\frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}=4.016\times 10^{-4}

Hence, the ratio of \frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}  in the cell is 4.016\times 10^{-4}

The ratio of the Sn2+ concentrations in the two half-cells is 0.00042.

The cell reaction is; Sn(s) ------> Sn^2+(aq) + 2e

Using the Nernst equation;

Ecell = E°cell - 0.0592/n log Q

Ecell = 0.10 V

n = 2

Q = ?

E°cell = 0.00 V

0.10 V = 0.00 V - 0.0592/2 log Q

-(0.10 V × 2/0.0592) = log Q

-3.378 = log Q

Q = 0.00042

Answer:

The i component of F is 7963.2 x 10⁻¹⁹ N.

The j component of F is 4561.2 x 10⁻¹⁹ N.

The k component of F is -151.2 x 10⁻¹⁹ N.

Magnitude of force is 9178.05 x 10⁻¹⁹ N.

Magnitude of force in presence of electric field is 27483.91 x 10⁻¹⁹ N.

Explanation:

Lorentz Force is defined as the force experience by charge particle in presence of electric and magnetic field. It is given by :

F = q ( v x B + E )     .....(1)

Here F is force, v is velocity of particle , B is magnetic field and E is electric field all are in vector notation and q is charge of the particle.

In first case, given :

E = 0

q = 24 x 10⁻¹⁹ C

v = [(-3.68)i + (6.35)j + (-1.91)k] x 105 m/s

B = [(-3.05)i + (5.28)j + (-1.09)k] T

Force, F = q x (v x B )

Substitute the suitable values in the above equation.

F = 24\times10^{-19}\times \left[\begin{array}{ccc}i&j&k\\-3.68&6.35&-1.91\\-3.05&5.28&-1.09\end{array}\right]\times105

F = 24 x 10⁻¹⁹ x { [6.35 x (-1.09) - (-1.91) x (5.28)] i - [(-3.68) x (-1.09) - (-1.91) x (-3.05)] j + [(-3.68) x (5.28) - 6.35 x (-3.05) ] k } x 105

F = 24 x 10⁻¹⁹ x 105 x [(3.16)i + (1.81)j + (-0.06)k] N   ....(2)

The i component of F is 7963.2 x 10⁻¹⁹ N.

The j component of F is 4561.2 x 10⁻¹⁹ N.

The k component of F is -151.2 x 10⁻¹⁹ N.

Magnitude of force, F =\sqrt{(7963.2\times10^{-19}) ^{2} +(4561.2\times10^{-19}) ^{2} +(-151.2\times10^{-19}) ^{2} }

F = 9178.05 x 10⁻¹⁹ N

In this case , E is also present and is equal to [(5.94)i + (-3.97)j + (-5.55)k] x 105 N/C. From equation (1) and (2) ;

F = 24 x 10⁻¹⁹ x [(3.16)i + (1.81)j + (-0.06)k + (5.94)i + (-3.97)j + (-5.55)k] x 105

F = 24 x 10⁻¹⁹ x 105 x [(9.1)i + (-2.16)j + (-5.61)k]

Magnitude of force, F = 24\times10^{-19}\times105\times\sqrt{9.1^{2}+(-2.16^{2})+(-5.61^{2}) }

F = 27483.9 x 10⁻¹⁹ N