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dusya [7]
2 years ago
13

An ordinary flashlight battery has a potential difference of 1.2 V between its positive and negative terminals. How much work mu

st you do to transport an electron from the positive terminal to negative terminal
Physics
1 answer:
Maru [420]2 years ago
4 0

The work done to transport an electron from the positive to the negative terminal is 1.92×10⁻¹⁹ J.

Given:

Potential difference, V = 1.2 V

Charge on an electron, e = 1.6 × 10⁻¹⁹ C

Calculation:

We know that the work done to transport an electron from the positive to the negative terminal is given as:

W.D = (Charge on electron)×(Potential difference)

       = e × V

       = (1.6 × 10⁻¹⁹ C)×(1.2 V)

       = 1.92 × 10⁻¹⁹ J

Therefore, the work done in bringing the charge from the positive terminal to the negative terminal is 1.92 × 10⁻¹⁹ J.

Learn more about work done on a charge here:

<u>brainly.com/question/13946889</u>

#SPJ4

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Semenov [28]

Answer:

f_n=3.75N

Explanation:

From the question we are told that:

Frictional force F=0.150N

Coefficient of kinetic friction \mu=0.04

Generally the equation for Normal for is mathematically given by

 f_n=\frac{F}{\mu}

Therefore

 f_n=\frac{0.150}{0.04}

 f_n=3.75N

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3 years ago
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Sholpan [36]

Answer:

If one plays ice hockey he should must have necessary equipment to support his sport. These equipment include Ice Skates,Helmet with Cage and Mouth-guard:, Hockey stick, Hockey pants,Hockey gloves, shoulder pads, elbow pads, Shin Guard:Neck guard and Jockstrap (men) or Pelvic protector (women)

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3 years ago
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gayaneshka [121]

Answer:

food

Explanation:

did you get a chance to look at the maximum number of devices allowed by

4 0
3 years ago
already did the work, i just need someone to see if i did the tangent line right? the line has to touch the point 0.6! thank you
Dmitrij [34]

The tangent looks good.

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5 0
3 years ago
Two cylindrical rods, one copper and the other iron, are identical in lengths and cross-sectional areas. They are joined, end to
Pie

Answer:

Vc = 2.41 v

Explanation:

voltage (v) = 16 v

find the voltage between the ends of the copper rods .

applying the voltage divider theorem

Vc = V x (\frac{Rc}{Rc + Ri})

where

  • Rc = resistance of copper = \frac{ρl}{a}  (l = length , a = area, ρ = resistivity of copper)
  • Ri = resistance of iron = \frac{ρ₀l}{a}  (l = length , a = area, ρ₀ = resistivity of copper)

Vc =  V x (\frac{\frac{ρl}{a}}{\frac{ρl}{a} + \frac{ρ₀l}{a}})

Vc = V x (\frac{ρ x (\frac{l}{a})}{(ρ + ρ₀) x (\frac{l}{a})})

Vc = V x (\frac{ρ}{ρ + ρ₀})

where

  • ρ = resistivity of copper = 1.72 x 10^{-8} ohm.meter
  • ρ₀ = resistivity of iron = 9.71 x 10^{-8} ohm.meter

Vc = 16 x (\frac{1.72 x 10^{-8}}{1.72 x 10^{-8} + 9.71 x 10^{-8}})

Vc = 2.41 v

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3 years ago
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