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Alja [10]
4 years ago
7

Static discharge differs from electric current because static discharge is described by which of the following?

Physics
2 answers:
inessss [21]4 years ago
6 0
<h2>Answer: C. Lasts only a fraction of a second</h2>

Explanation:

got it right in my exam

Bond [772]4 years ago
4 0
<span>Static discharge differs from electric current because static discharge is described by C. lasting for only a fraction of a second.
This is a very short period of time, and electric current lasts longer than a static discharge does. 
</span>
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3 years ago
What factors affect the speed of water waves
Aneli [31]
Hey there,

Your question states: What factors affect the speed of water waves
Let's get one thing out the way, (wavelength) does NOT affect the the speed of water. If anything, it would be how high the wavelength's are. The higher the wavelengths are, the more that it would affect the speed, because there very high, but if it were to go longer on the width side, that would increase the speed, but that's not the case. Your correct answer would be (higher wavelength).

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6 0
3 years ago
A light bulb operating at a dc voltage of 120 V has a power rating of 60 W. How much current is flowing through this bulb
densk [106]
  • Voltage=V=120V
  • Power=P=60W

  • Current=I

\\ \rm\rightarrowtail I=\dfrac{P}{V}=\dfrac{60}{120}=\dfrac{1}{2}=0.5A

8 0
2 years ago
Will a substance that contains an o-h group dissolve in water?
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Usually, it increases the solubility in water.
4 0
3 years ago
A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary
Snezhnost [94]

Explanation:

Let N_p\ and\ N_s are the number of turns in primary and secondary coil of the transformer such that,

\dfrac{N_p}{N_s}=\dfrac{1}{3}

A resistor R connected to the secondary dissipates a power P_s=100\ W

For a transformer, \dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}

V_s=(\dfrac{N_s}{N_p})V_p

V_s=3V_p...............(1)

The power dissipated through the secondary coil is :

P_s=\dfrac{V_s^2}{R}

100\ W=\dfrac{V_s^2}{R}

V_p^2=\dfrac{100R}{9}.............(2)

Let N_p'\ and\ N_s' are the new number of turns in primary and secondary coil of the transformer such that,

\dfrac{N_p'}{N_s'}=\dfrac{1}{24}

New voltage is :

V_s'=(\dfrac{N_s'}{N_p'})V_p'

V_s'=24V_p...............(3)

So, new power dissipated is P_s'

P_s'=\dfrac{V_s'^2}{R}

P_s'=\dfrac{(24V_p)^2}{R}

P_s'=24^2\times \dfrac{(V_p)^2}{R}

P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}

P_s'=6400\ Watts

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0
3 years ago
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