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Westkost [7]
3 years ago
8

Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B​

Physics
1 answer:
andreyandreev [35.5K]3 years ago
8 0

Explanation:

Let \textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}} and \textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}

The sum of the two vectors is

\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}

= 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}

The difference between the two vectors can be written as

\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}

= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}

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ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami
yuradex [85]

Answer : The values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 62.2kJ,132.67J/K\text{ and }22.66kJ respectively.

Explanation :

The given balanced chemical reaction is,

2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{product}

\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]

\Delta H^o=62.2kJ=62200J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{product}

\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]

\Delta S^o=132.67J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 298 K.

\Delta G^o=(62200J)-(298K\times 132.67J/K)

\Delta G^o=22664.34J=22.66kJ

Therefore, the values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 62.2kJ,132.67J/K\text{ and }22.66kJ respectively.

6 0
3 years ago
A swimming pool is 25.0 ft. long, 18.5 ft. wide, and 9.0 ft. deep. When filled, the water level is 7.0 inches from the top. Disi
GarryVolchara [31]

Answer:

1.9841256 kg

Explanation:

Given;

Length of the swimming pool = 25.0 ft = 7.62 m   ( 1 ft = 0.3048 m )

Width of the swimming pool = 18.5 ft =  5.64 m

Depth of the pool = 9.0 ft =

Total depth of the water in the pool when filled = 9 ft - 7 inches = 2.56 m

now,

Volume of the water in the pool = Length × Width × Depth

or

Volume of the water in the pool = 7.62 × 5.64 × 2.56 = 110.2292 m³

also,

1 m³ = 1000 L

thus,

110.2292 m³ = 110229.2 L

also it is given that 18 mg of Cl is added to 1 liter of water

therefore,

In 110229.2 L of water Cl added will be = 110229.2 × 18 = 1984125.6 mg

or

= 1.9841256 kg

8 0
3 years ago
How are the oscillating magnetic and electric fields of an electromagnetic wave positioned relative to each other?​
Pepsi [2]

Answer:

Electromagnetic waves consist of both electric and magnetic field waves. These waves oscillate in perpendicular planes with respect to each other, and are in phase. The creation of all electromagnetic waves begins with an oscillating charged particle, which creates oscillating electric and magnetic fields.

Explanation:

7 0
2 years ago
The magnetic domains in an iron nail are randomized unless it is placed in a magnetic field.
UNO [17]

Answer:

True plz thank me this is the answer

3 0
2 years ago
A student sits in a chair that can spin without friction. The student has her hands outstretched and starts rotating at 1.9 rev/
Airida [17]

Answer:

the resulting angular speed after she pulls her hand inwards in (rad/s)  is  27.02 rad/s

Explanation:

Given that :

the initial angular speed \omega_1 = 1.9 \ \ rev/s

Initial rotational inertia I_1 = 12.00 \ \ kg.m^2

Final angular speed \omega_2 = ???

Final rotational inertia I_2 = 5.3 \ \ kg.m^2

According to conservation of momentum :

Initial momentum = final momentum

I_1 \omega_1 = I_2 \omega_2

\omega_2 = \frac{I_1 \omega_1}{I_2}

\omega_2 = \frac{12.00*1.9}{5.3}

\omega_2 = 4,3 rev/s

To rad/s ; we have:

1  \   rev/s = 2 \pi \ \  rad/s

\omega_2 = 4.3 * 2 \pi \ \  rad/s

\omega_2 = 27.02  \ \ rad/s

Therefore the resulting angular speed after she pulls her hand inwards in (rad/s) = 27.02 rad/s

8 0
3 years ago
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