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8_murik_8 [283]
3 years ago
14

a portable DVD player requires six 1.5-volt batteries to operate.if the resistance in the circuitis 7.2 ohms. what's the current

?
Physics
1 answer:
kupik [55]3 years ago
7 0

Hello!

A portable DVD player requires six 1.5-volt batteries to operate. If the resistance in the circuit is 7.2 ohms, what is the current?

Data:

We have six 1.5 volt batteries, soon, your tension (V)

V = 1.5 * 6 = 9 Volts

R (resistance) = 7.2 ohms

I (Current intensity) = ? (in Amps)

Solving (Ohm's First Law):

I = \dfrac{V}{R}

I = \dfrac{9}{7.2}

\boxed{\boxed{I= 1.25\:A}}\:\:\:\:\:\:\bf\blue{\checkmark}

Answer:  

The current is 1.25 amps

________________________

\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

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34kurt

Poder = (resistencia) x (corrente)²

Poder = (10 ohms) x (5 A)²

<em>Poder = 250 watts </em>(250 Joule por segundo)

2 horas = 7,200 segundos

Energia = (250 joule/seg) x (7,200 seg)

<em>Energia = 1,800,000 Joules</em>

7 0
3 years ago
A wave is produced in a rope. The wave has a speed of 33 m/s and a frequency of 22 Hz. What wavelength is produced? 0. 67 m 0. 7
vekshin1

Answer:

Wavelength = <u>1.5 m</u>

Explanation:

The formula for waves in terms of wavelength, speed and frequency is:

Speed (v) = Frequency (f) × Wavelength (λ)

33 = 22 × λ

33 = 22λ

λ = \frac{33}{22}

So, λ = 1.5 m

4 0
2 years ago
An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea
melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, v=1.37\times10^7\ \rm m/s

Explanation:

<u>Given:</u>

  • Linear charge density of the ring=0.1\ \rm \mu C/m
  • Radius of the ring R=0.2 m
  • Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C

Potential due the ring at a distance x from the centre of the rings is given by

V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V

Let\Delta U be the change in potential Energy given by

\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s

So the electron will be moving with v=1.37\times10^7\ \rm m/s

5 0
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nadezda [96]
Answer: 11,100 ft/s^2

1) Constant acceleration=> uniformly accelerated motion.

2)  Formula for uniformly accelerated motion:

Vf = Vo + at

3) Data:

Vo = 1,100 ft/s
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t = 10.0 s

4) Solution:

Vf = 1,100 ft/s + 1,000 ft/s^2 * 10.0 s = 1,100 ft/s + 10,000 ft/s

Vf = 11,100 ft/s

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Answer:

A

Explanation:

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