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Ipatiy [6.2K]
3 years ago
10

What is the main reason why many nuclear power plants are located near bodies of water?

Physics
2 answers:
PilotLPTM [1.2K]3 years ago
6 0
The answer is C but man if you have leak or a meltdown good luck to anyone downstream.

Alex17521 [72]3 years ago
3 0

What is the main reason why many nuclear power plants are located near bodies of water?

the answer is c i just got done doing the test!!

your welcome

Good luck

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The eye is actually a multiple-lens system, but we can approximate it with a single-lens system for most of our purposes. When t
hram777 [196]

Explanation:

Given that,

The optical power of the equivalent single lens is 45.4 diopters.

(a) The relationship between the focal length and the focal length is given by:

f=\dfrac{1}{P}

f=\dfrac{1}{45.4}

f = 0.022 m

or

f = 2.2 cm

(b) We need to find how far in front of the retina is this "equivalent lens" located. It is given by using lens formula as :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Here, u = infinity

\dfrac{1}{v}=\dfrac{1}{2.2}

v = 2.2 cm

So, at 2.2 cm in front of the retina is this "equivalent lens" located.

Hence, this is the required solution.

5 0
3 years ago
A 120 Ω resistor, a 60 Ω resistor, and a 40 Ω resistor are connected in parallel and placed across a potential difference of 12.
Westkost [7]

Answer:

The equivalent resistance of the parallel circuit would be 20 Ω

Explanation:

To calculate the resistance of resistors connected in parallel, the formula to be used is

1/R = 1/R₁ + R₂ + R₃ + R₄...

1/R = 1/120 + 1/60 + 1/40

1/R = (1 + 2 + 3)/120

1/R = 6/120

1/R = 1/20 Ω

This can be rewritten or cross-multiplied to be

R × 1 = 20 × 1

R = 20 Ω

The equivalent resistance (R) would then be 20 Ω

7 0
3 years ago
A 4kg and 5kg bodies moving on a frictionless horizontal surface at a velocity of ( -6i )m/s and ( +3 )m/s respectively. Collide
Vladimir [108]

Answer:

4 kg → +4 m/s

5 kg → -5 m/s

Explanation:

The law of conservation of momentum states that:

  • m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
  • left side → velocities before collision
  • right side → velocities after collision

You'll notice that we have two missing variables: v₁' & v₂'. Assuming this is a perfectly elastic collision, we can use the conservation of kinetic energy to set the initial and final velocities of the individual bodies equal to each other.

  • v₁ + v₁' = v₂ + v₂'  

Let's substitute all known variables into the first equation.

  • (4)(-6) + (5)(3) = (4)v₁' + (5)v₂'
  • -24 + 15 = 4v₁' + 5v₂'
  • -9 = 4v₁' + 5v₂'  

Let's substitute the known variables into the second equation.

  • (-6) + v₁' = (3) + v₂'
  • -9 = -v₁' + v₂'
  • 9 = v₁' - v₂'  

Now we have a system of equations where we can solve for v₁ and v₂.

  • -9 = 4v₁' + 5v₂'
  • 9 = v₁' - v₂'  

Use the elimination method and multiply the bottom equation by -4.

  • -9 = 4v₁' + 5v₂'
  • -36 = -4v₁' + 4v₂'

Add the equations together.

  • -45 = 9v₂'
  • -5 = v₂'

<u>The final velocity of the second body (5 kg) is -5 m/s</u>. Substitute this value into one of the equations in the system to find v₁.  

  • 9 = v₁' - v₂'
  • 9 = v₁' - (-5)
  • 9 = v₁' + 5
  • 4 = v₁'

<u>The final velocity of the first body (4 kg) is 4 m/s.</u>

<u></u>

We can verify our answer by making sure that the law of conservation of momentum is followed.

  • m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
  • (4)(-6) + (5)(3) = (4)(4) + (5)(-5)
  • -24 + 15 = 16 - 25
  • -9 = -9

The combined momentum of the bodies before the collision is equal to the combined momentum of the bodies after the collision. [✓]

3 0
2 years ago
A plane traveled a distance of 4,200 km for 3.5 hours. What was the speed of the plane?
bija089 [108]

Answer:

1200km per hour

Explanation:

3 0
3 years ago
Read 2 more answers
A certain superconducting magnet in the form of a solenoid of length 0.24 m can generate a magnetic field of 7.0 T in its core w
tatiyna

Answer:

N=14854.5turns

Explanation:

Given data

Length L=0.24 m

Magnetic field B=7.0 T

Current I=90 A

To find

Number of turns N

Solution

As we know that magnetic field is given as:

B=u_{o}I\frac{N}{L}

Rearrange the equation and solve for N

So

B=u_{o}I\frac{N}{L}\\\frac{B}{u_{o}I} =\frac{N}{L} \\N=\frac{LB}{u_{o}I}\\N=\frac{(0.24m)(7.0T)}{4\pi *10^{-7}*(90A)} \\N=14854.5turns

5 0
3 years ago
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