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SashulF [63]
2 years ago
7

A current of 0.050 amps flows in a circuit element with a resistance of 25 ohms what is the voltage drop across the circuit elem

ent. (use ohms law V=IR)

Physics
1 answer:
katen-ka-za [31]2 years ago
3 0

The voltage drop across the circuit, given that 0.05 A flows in the circuit is 1.25 V

<h3>Ohm's law</h3>

This law states as follows:

V = IR

Where

  • V is the potential difference
  • I is the current
  • R is the resistor

<h3>How to determine the voltage</h3>

From the question given above, the following data were obtained:

  • Current (I) = 0.05 A
  • Resistance (R) =25 ohms
  • Voltage (V) =?

V = IR

V = 0.05 × 25

V = 1.25 V

Therefore, the voltage drop across the circuit is 1.25 V

<h3 />

Learn more about Ohm's law:

brainly.com/question/796939

#SPJ1

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Given that,

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As 10Kg > 5Kg , the 10 kg mass accelerates down and the 5kg mass accelerates up, let it be a. Let the tension in the cable be T.

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Substituting them back in the equation we get,

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Which is a disadvantage of using wind as an energy source?
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Answer:

it produces

Explanation:

a lot of waste

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during a lunar mission, it is necessary to increase the speed of a spacecraft by 2.2 m/s when it is moving at 400 m/s relative t
wel

The initial mass fraction of the spacecraft that must be burned and ejected to achieve an increase in speed is 0,00219 m/s

<h3>What fraction of the initial mass of the spacecraft?</h3>

Increase the speed: Vf-Vi = 2.2 m/s

Speed of aircraft: Vr = 400 m/s

Speed of ejected products: Vrel = 1000 m/s

The answer is:

V_f - V_i = V_{rel} log_{e} (\frac{mi}{mf})\\\\2.2 = 1000 log_{e} (\frac{mi}{mf})\\\\ log_{e} (\frac{mi}{mf} ) = \frac{2.2}{1000} \\\\ log_{e} (\frac{mi}{mf} ) = 0.0022

\frac{mi}{mf} = e^{0,0022} \\\\\frac{mf}{mi} = e^{-0,0022} \\\\\frac{mi-mf}{mi} = 1 - e^{-0,0022} = 0,00219

So, the initial mass fraction of the spacecraft that must be burned and ejected to achieve an increase in speed is 0,00219 m/s

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3 0
1 year ago
The potential difference between the plates of a capacitor is 145 V. Midway between the plates, a proton and an electron are rel
aniked [119]

Answer:

= 2.52 x 10^ 6 m/s        

Explanation:

The force that acts on charged particles between capacitor plates =

F = (q) (Δv)  ÷ d

Here,  d = distance between the two plates

          q = charge of the charged particle

         Δv = voltage

Normally, the force that makes both proton and electron released from rest, giving the charge acceleration is F=m X a. where m= mass and a = acceleration

Poting this equation with the first one, we have:

m X a =  (q) (Δv)  ÷ d

So, the acceleration of a proton when moving towards a negatively charged plate is

a = (q) (Δv)  ÷ (d) (m) {proton}

Likewise, the acceleration of an electron when moving towards a positively charged plate is

a = (q) (Δv)  ÷ (d) (m) {electron}

Dividing the proton acceleration formula by the electron acceleration formula we have:

a (proton) / a (electron) = m (proton) / m(electron)

inserting equation of motion to get distance, s

s = ut + 1/2 at^2

recall that electron travel distance, d/2

d/2 = 1/2 at^2

making t the subject of the formula

we have, t =√(d ÷ a(electron))

The distance of proton:

d/2 =  ut + 1/2 at^2 [proton}

put d/2 =  ut + 1/2 at^2 [proton} into t =√(d ÷ a(electron))

Initial speed, ui = √(d ÷ a(electron)) = (d/2) - (1/2) x (d) (a(proton) + a(electron))

since acceleration wasn't given in the question, lets use mass(elect

ron)  ÷ mass(proton) rather than use (a(proton) + a(electron))

Therefore, intial speed= 1/2√((e X Δv) ÷ m(electron)) (1- m(electron)/ m(proton))

   Note, e = 1.60 x 10^-19

           m(electron) = 9.11 X 10^-31

            m(proton) = 1.67  X 10^-27

Input these values into the formula above, initial speed, UI =  

           = 2.52 x 10^ 6 m/s          

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3 years ago
It has been argued that power plants should make use of off-peak hours to generate mechanical energy and store it until it is ne
sdas [7]

Answer:

Explanation:

90 rpm = 90 / 60 rps

= 1.5 rps

= 1.5 x 2π rad /s

angular velocity of flywheel

ω = 3π rad /s

Let I be the moment of inertia of flywheel

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=2 x  10⁷ / (3π)²

= 2.2538 x 10⁵ kg m²

Let radius of wheel be R

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= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .

1/2 πR⁴ x t x d = 2.2538 x 10⁵

R⁴ = 2 x 2.2538 x 10⁵ / πt d

= 4.5076 x 10⁵ / 3.14 x .1 x 7800

= 184

R= 3.683 m .

diameter = 7.366 m .

b ) centripetal accn required

= ω² R

= 9π² x 3.683

= 326.816 m /s²

3 0
3 years ago
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