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antoniya [11.8K]
3 years ago
6

Find the current passing through each of the 3 resistors connected parallel to each other as shown in the figure (i1, i2, I3). S

how your actions clearly.

Physics
1 answer:
Thepotemich [5.8K]3 years ago
3 0

Answer:

I1 = ε/R1

I2 = ε/R2

I3 = ε/R3

Explanation:

From the image, we see that the resistors are connected in parallel. This means that the voltage passing through them is the same.

Now, formula for current is; I = V/R

In this case, V which is voltage is denoted by ε.

Thus;

I1 = ε/R1

I2 = ε/R2

I3 = ε/R3

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Answer

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5 0
3 years ago
Whats this symbol?<br> ∑
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4 0
3 years ago
Suppose you are an astronaut and you have been stationed on a distant planet. You would like to find the acceleration due to the
Maksim231197 [3]

Answer:

<h3> 1.40625m/s²</h3>

Explanation:

Using the equation of motion expressed as v = u+gt where;

v is the  final velocity of the ball

u is the initial velocity

g is the acceleration due to gravity

t is the time taken

Given

u = 9m/s

v = 0m/s

t = 6.4s

Required

acceleration due to gravity g

Since the rock is thrown up, g will be a negative value.

v = u+(-g)t

0 = 9-6.4g

-9 = -6.4g

6.4g = 9

divide both sides by 6.4

6.4g/6.4 = 9/6.4

g = 1.40625m/s²

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6 0
3 years ago
Why is the current atomic model called the "electron could model"
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8 0
3 years ago
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
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