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3 years ago

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Two long, straight, parallel wires are carrying a current I in the same direction. Select all that apply to the force felt by on

e wire due to the other.
a. It is parallel to the current in the other wire.
b. It is perpendicular to the field produced by the other wire.
c. It is perpendicular to the current in the other wire.
d. It is parallel to the field produced by the other wire.

1 answer:

Fiesta28 [93]3 years ago

4 0

Answer: Option C.

It is perpendicular to the field produced by the other wire

Explanation

It is perpendicular to the field produced by the other wire because the force between the two parallel line is attractive in nature because the current is flowing in the same direction.

The two parallel wires carrying current in the same direction attract each other and their magnetic field will be attracted to each other.

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If 20 beats are produced within one second, which of the following frequencies could possibly be held by two sound waves traveli

NeTakaya

Answer:

D. 22 Hz and 42 Hz

Explanation:

When two waves with different frequency travelling in the same medium meet each other, they produce an interference pattern called beat.
<em><u>The frequency of the beat produced is equivalent to </u></em><em><u>the difference between the individual frequencies of the two waves involved.</u></em>
<em><u>Therefore; in this case since the frequency of the beat is 20 Hz, that is from 20 beats per second.</u></em>
We need to find a pair from the choices whose frequency difference is 20 Hz.
This happens to be choice D. 22 Hz and 42 Hz, that is 42 Hz - 22 Hz = 20 Hz

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3 years ago

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What is the relationship between mass gravity and weight?

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Fg=m•g || IE: Weight = mass x gravity
Therefore, the relationship are as follows:
mass and gravity are inversely proportional
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weight and gravity are directly proportional

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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.

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Maru [420]

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Consider the expression below. Assume m is an integer. 6m(2m + 18) Enter an expression in the box that uses the variable m and m

Anika [276]

6x2=12m
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