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katovenus [111]
3 years ago
14

Some superconductors are capable of carrying a very large quantity of current. If the measured current is 1.00 ´ 105 A, how many

electrons are moving through the superconductor per second? (e = 1.60 ´ 10-19 C)
Physics
1 answer:
Zielflug [23.3K]3 years ago
5 0

Answer:

The 6.25 \times 10^{23} electrons are moving through the superconductor per second.

Explanation:

Given :

Current I = 1 \times 10^{5} A

Charge of electron e = 1.6 \times 10^{-19} C

Time t = 1 sec

From the formula of current,

Current is the number of charges flowing per unit time.

   I = \frac{ne}{t}

Where n = number of charges means in our case number of electrons

   n = \frac{It}{e}

   n = \frac{1 \times 10^{5} }{1.6 \times 10^{-19} }

   n = 6.25 \times 10^{23}

Therefore, 6.25 \times 10^{23} electrons are moving through the superconductor per second.

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Answer:

Time = 2758.62 seconds

Explanation:

Given the following data;

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Time = distance/speed

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Therefore, it will take the pilot 2758.62 seconds to reach the airport.

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2 years ago
When you stand with the wind blowing on your back, the low pressure center is
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The velocity of a particle is described
Dmitry_Shevchenko [17]

Answer:

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4 0
3 years ago
A 2 kg ice cube is released from point A and slides on a frictionless track as shown in the figure. Determine the speed of the c
kherson [118]

According to the conservation of energy

  • Potential energy at any given instance is equal to the Kinetic energy as energy can neither be created nor be destroyed

Mass is 2kg=m

#A

h=5m

  • PE=mgh

PE

  • 2(9.8)(5)
  • 10(9.8)
  • 98J

Now

  • KE=98J
  • 1/2mv²=98J
  • 1/2×2v²=98J
  • v²=98J
  • v=√98
  • v=9.4m/s

#B

h=3.2m

PE:-

  • 2(3.2)(9.8)
  • 6.4(9.8)
  • 62.7J

Now

  • KE=62.7J
  • 1/2mv²=62.7
  • v²=62.7
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#C

h=2m

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  • 2(2)(9.8)
  • 4(9.8)
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Now

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8 0
2 years ago
It is generally a good idea to gain an understanding of the "size" of units. Consider the objects listed and calculate the magni
Alex

Answer:

Explanation:

The standard unit of momentum Is kgm/s

And momentum is the product of mass and velocity

Then,

Momentum = mass × velocity

P = mv

Where

m is mass in kg

v is velocity in m/s

Now,

Given that,

M = 31.8mg

1mg = 10^-3g

So, m = 31.8 × 10^-3g

Also, 1000g = 1kg → 1g = 10^-3 kg

So,

m = 31.8 × 10^-3 × 10^-3kg

m = 31.8 × 10^-6 kg

v = 2.53mi/hr

1mile = 1609.34 m

1 hour = 3600seconds

Then,

v = 2.53 × 1609.34m/ 3600s

v = 1.13 m/s

So, momentum is given as

P = mv

P = 31.8 × 10^-6 × 1.13

P = 3.597 × 10^-5 kgm/s

5 0
3 years ago
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