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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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Answer:

B = 0.8 T

Explanation:

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The loop may be rotated about an axis that passes through the center and lies in the plane of the loop.

When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 Nm.

We need to find the magnitude of the uniform magnetic field exerting this torque on the loop. Torque acting on the loop is given by :

\tau=NIAB\sin\theta

B is magnetic field

B=\dfrac{\tau}{NIA\sin\theta}\\\\B=\dfrac{1.8}{1\times \pi \times (0.75)^2\times 3\times \sin(25)}\\\\B=0.8\ T

So, the magnitude of the uniform magnetic field exerting this torque on the loop is 0.8 T.

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A

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2 years ago
With respect to the earth, object 1 is moving at speed 0.80 c to the right. Object 2 is moving in the same direction at speed 0.
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Answer:

0.976 c

Explanation:

v_{1e} = velocity of object 1 relative to earth = 0.80 c

v_{21} = velocity of object 2 relative to object 1 = 0.80 c

v_{2e} = velocity of object 2 relative to earth

Velocity of object 2 relative to earth is given as

v_{2e}= \frac{v_{1e} + v_{21}}{1 + \frac{v_{1e}v_{21}}{c^{2}}}

v_{2e}= \frac{0.80 c + 0.80 c}{1 + \frac{(0.80c)(0.80c)}{c^{2}}}

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2 years ago
A object is located 16.0 cm in front of a converging lens whose focal length is 12.0 cm. A diverging lers ength of 10.0 cm is lo
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Answer:

first lens v = 48 cm

second lens v = -15.6 cm

magnification =  1.67

final image is virtual

and final image is upright

Explanation:

given data

distance = 16 cm

focal length f1 = 12 cm

focal length f2 = 10.0 cm

to find out

location of the final image and magnification and Type of image

solution

we apply here lens formula that is

1/f = 1/v + 1/u     .....................1

put here all value and find v  for 1st lens

1/12 = 1/v + 1/16

v = 48 cm

and  find v  for 2nd lens

here u = 20- 48 = -28

- 1/10 = 1/v - 1/28

v = -15.6 cm

and

magnification = first lens (v/u) ×  second lens ( v/u)

magnification =  (-15.6/-28) ×  ( 48/16)

magnification =  1.67

so here final image is virtual

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