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leonid [27]
2 years ago
12

A charge of 0.91 C is spread uniformly throughout a 25 cm rod of radius 4 mm. What are the volume and linear charge densities

Physics
1 answer:
Oxana [17]2 years ago
4 0

The volume of the rod is 1.26×10⁻⁵ m³, and the linear charge density of the rod is 3.64 C/m

<h3>What is volume?:</h3>

This is the product of the height of a solid object and its crossectional area.

The Volume of the rod is can be calculated using the formula below.

Note: A rod has the shape of a cylinder.

Formula:

  • V = πr²h............... Equation 1

Where:

  • V = Volume of the rod
  • r = radius of the rod
  • h = height of the rod.

From the question,

Given:

  • r = 4mm = 0.004 m
  • h = 25 cm = 0.25 m
  • π = 3.14

Substitute these values into equation 1

  • V = 3.14(0.004²)(0.25)
  • V = 1.26×10⁻⁵ m³

<h3>What is linear charge density:</h3>

This is the ratio of the charge on an object to the length of the object.

The linear charge density of the rod can be calculated using the formula below.

  • D = Q/h.................... Equation 2

Where:

  • D = Linear charge density of the rod
  • Q = Charge on the rod.
  • h = height or length of the rod

From the question

Given:

  • Q = 0.91 C
  • h = 25 cm = 0.25 m

Substitute these values into equation 2

  • D = 0.91/0.25
  • D = 3.64 C/m

Hence, The volume of the rod is 1.26×10⁻⁵ m³, and the linear charge density of the rod is 3.64 C/m

Learn more about charge density here: brainly.com/question/14568868

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Differences in ocean-surface height can be measured by _____.\
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<h3><u>Answer;</u></h3>

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<h3><u>Explanation;</u></h3>
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Answer:

T_A_v_g=9.918192559$^{\circ}C

Explanation:

The problem tell us that the temperature as function of time in downtown mathville is given by:

T(t)=10-5*sin(\frac{\pi}{12t})

The average temperature over a given interval can be calculated as:

T_a_v_g=\frac{T_o+T_f}{2}

Where:

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So, the initial temperature in this case, would be the temperature at noon, and the final temperature would be the temperature at midnight:

Therefore:

T_o=T(12)=10-5*sin(\frac{\pi}{12*12}) =9.890925575$^{\circ}C

T_f=T(24)=10-5*sin(\frac{\pi}{12*24}) =9.945459543$^{\circ}C

Hence, the average temperature between noon and midnight is:

T_A_v_g=\frac{9.890925575+9.945459543}{2}=9.918192559$^{\circ}C

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