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irina [24]
3 years ago
7

The potential difference across a piece of wire is 2.1 v the current in the wire is 0.30 a calculate the resistance of the wire

in ohms
Physics
1 answer:
IrinaVladis [17]3 years ago
8 0

Answer:

<h2>7Ω</h2>

Explanation:

Given:

V = 2.1 v

I = 0.30 A

R = Unknown

Solution:

R = V / I

= 2.1V / 0.30A

= 7Ω

Answer:

7Ω is the resistance of the wire in ohms

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Explanation A 5000 kg rocket is at rest in deep space. The rocket burns fuel pushing 10kg of exhaust gases rearward at 4000 m/s.
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Answer:

F = 4000 N

Explanation:

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 M v + m v' = 0

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now,

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F = \dfrac{m(v_i-v_f)}{\Delta t}

F = \dfrac{5000(8-0)}{10}

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A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left
Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

  • Length of the massless beam;L = 4.00m
  • Distance of support from the left end; x = 3.00m
  • First mass; m1 = 31.3 kg
  • Distance of beam from  the left end( m₁ is attached to ); x_1 = ?
  • Second mass; m_2 = 61.7 kg
  • Distance of beam from  the right of the support( m₂ is attached to ); x_1 = 0.273m

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, m_1g( x-x_1) = m_2gx_2

we divide both sides by g

m_1( x-x_1) = m_2x_2

Next, we make x_1, the subject of the formula

x_1 = x - [ \frac{m_2x_2}{m_1} ]

We substitute in our given values

x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]

x_1 = 3.00m - 0.538m

x_1 = 2.46m

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

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