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4 years ago

5

The conductors that carry the current to electrical devices and ? equipment are the heart of all electrical systems. There are a

ssociated ? whenever current flows through a conductor.

1 answer:

Aleksandr [31]4 years ago

7 0

Answer:

Utilization, effects

Explanation:

The conductors that carry the current to electrical devices and utilization equipment are the heart of all electrical systems. There are associated effects whenever current flows through a conductor.

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Compare to a low wavelength sound wave, a high wavelength sound wave Select all that apply Group of answer choices travels slowe

viktelen [127]

Answer:

B. has a smaller frequency

C. travels at the same speed

Explanation:

The wording of the question is a bit confusing, it should be short/long for wavelength and low/high for frequency. I assume low wavelength mean short wavelength.

All sound wave travel with the same velocity(343m/s) so wavelength doesn't influence its speed at all. It won't be faster or slower, it will have the same speed.

Velocity is a product of wavelength and frequency. So, a long-wavelength sound wave should have a lower frequency.

The option should be:

A. travels slower -->false

B. has a smaller frequency -->true

C. travels at the same speed --->true

D. has a higher frequency --->false

E. travels faster has the same frequency --->false

5 0

3 years ago

Example of energy transfer from potential to kinetic and back

Alex_Xolod [135]

Swimmer and Divers. The Potential energy is transferred into Kinetic energy, and allows the diver to submerge into the water. The Kinetic energy then allows the diver to submerge and dive into the water. Potential energy however, is needed to allow the diver to get back out of the water after diving to get up and go and dive again, and then the Kinetic energy is transferred back to Potential energy to repeat the process.

Hope :) -Emilie Xo this is right and it helps! Xo

5 0

3 years ago

Net force is the sum of all forces acting on an object. For example, in a tug of war, when one team is pulling the rope with a f

melisa1 [442]

Answer:

If there are equal forces in both directions and there is no motion, the net force is 0 Newtons. This is because you'd be subtracting 100 from 100 which just equals 0.

5 0

3 years ago

A man 2 meters tall walks at the rate of 2 meters per second toward a streetlight that's 5 meters above the ground. At what rate

GaryK [48]

Answer:

It changes at a rate of 4/3 meter per second

Explanation:

In the given figure below we have

$\Delta OBD\simeq \Delta ABC\\\\\therefore \frac{5}{X+Y}=\frac{2}{Y}\\\\$

Solving for Y given $X=2m/s$ we get

$\frac{5}{2+Y}=\frac{2}{Y}\\\\5Y=4+2Y\\\\Y=\frac{4}{3}m/s$

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