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abruzzese [7]
3 years ago
5

Find the voltage drop (in volt) along a 93.4 meter long 10 gauge copper wire carrying acurrent of 72.5 A. The diameter of a 10 g

auge wire is 2.5882 millimeters.
Physics
1 answer:
Sophie [7]3 years ago
3 0

Answer: 5.41 V

Explanation:in order to explain this result we have to use the Ohm law given by:

ΔV=R*I where R is the resistance which is equal R= ρ*L/A . ρ is the resistivity, L the length of the wire and A is the cross section. I is the current.

Then we have

ΔV=ρ*L*I/A= 1.68 * 10^-8 Ωm*93.4 m*72.5A/2.1* 10^-5 m^2=5.41 V

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murzikaleks [220]

Answer:

160 Nm

Explanation:

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Frick and Frack are standing back-to-back, leaning on each other, but not moving. If Frick weighs
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Answer:

Frick is pushing harder

Explanation:

if Frack weighs more and he was pushing harder they would be moving, but if Frick pushes harder then they wont move

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Grinding pepper is a physical change
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In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the
MatroZZZ [7]

Answer:

The wavelength is \lambda_2 =  534 *10^{-9} \ m

Explanation:

From the question we are told that

   The wavelength of the first light is  \lambda _ 1 =  587 \ nm

    The order of the first light that is being considered is  m_1  =  10

     The order of the second light that is being considered is  m_2  =  11

Generally the distance between the fringes for the first light is mathematically represented as

      y_1 =  \frac{ m_1  * \lambda_1 *  D}{d}

 Here  D is the distance from the screen

 and    d  is the distance of separation of the slit.

      For the second light the distance between the fringes is  mathematically represented as

         y_2 =  \frac{ m_2  * \lambda_2 *  D}{d}

Now given that both of the light are passed through the same double slit

       \frac{y_1}{y_2}  =  \frac{\frac{m_1 *  \lambda_1 * D}{d}  }{\frac{m_2 *  \lambda_2 * D}{d}  } = 1

=>    \frac{ m_1 *  \lambda _1  }{ m_2  *  \lambda_2} =  1

=>     \lambda_2 =  \frac{m_1 *  \lambda_1}{m_2}

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=>   \lambda_2 =  534 *10^{-9} \ m

4 0
3 years ago
A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out
anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

KE_{initial} + PE_{initial} = KE_{final} + PE_{final}

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

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v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

mv = (m + m')v'

65.0\times 9.89 = (65.0 + 20.0)v'

v' = 7.56\ m/s

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h = \frac{1}{2}gt^{2}

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s

Now, the horizontal distance is given by:

x = v't = 7.56\times 0.638 = 4.823\ m

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3 years ago
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