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hjlf
2 years ago
14

Aluminium,metal,glass,and paper:

Physics
1 answer:
Travka [436]2 years ago
7 0

Aluminum metal is a good example of conductor of electricity while glass and paper are good example of insulators.

<h3>What is a conductor?</h3>

A conductor is material that allows easy passage of electric current through the.

<h3>Examples of conductors</h3>

All metals are good examples conductors. such as aluminum, iron, silver, etc.

<h3>What is a non conductor?</h3>

A  non conductor or insulator is a material which does not allow easy passage of electric current through them.

<h3>Examples of insulators</h3>
  • glass
  • paper
  • wood, etc

Thus, aluminum metal is a good example of conductor of electricity while glass and paper are good example of insulators.

The complete question is below:

aluminium,metal,glass,and paper, are examples of what?

Learn more about conductors here: brainly.com/question/24154868

#SPJ1

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A wire of radius R has a current I uniformly distributed across its cross-sectional area. Ampere's law is used with a concentric
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Answer:

Please refer to the figure.

Explanation:

The crucial point here is to calculate the enclosed current. If the current I is flowing through the whole cross-sectional area of the wire, the current density is

J = \frac{I}{\pi R^2}

The current density is constant for different parts of the wire. This idea is similar to that of the density of a glass of water is equal to the density of a whole bucket of water.

So,

J = \frac{I}{\pi R^2} = \frac{I_{enc}}{\pi r^2}\\I_{enc} = \frac{Ir^2}{R^2}

This enclosed current is now to be used in Ampere’s Law.

\mu_o I_{enc} = \int {B} \, dl

Here, \int \, dl represents the circular path of radius r. So we can replace the integral with the circumference of the path, 2\pi r.

As a result, the magnetic field is

B = \frac{\mu_0}{2\pi}\frac{Ir}{R^2}

5 0
3 years ago
A Carnot air conditioner operates between an indoor temperature of 20°C and an outdoor temperature of 39°C. How much energy does
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Answer:

D. 130 J

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The coefficient of performance for a machine that is being used to cool, is given by:

COP=\frac{Q_C}{W}=\frac{T_C}{T_H-T_C}

Here Q_C  is the heat removed from the cold reservoir, W is the work required, that is, the energy required to remove the heat from the interior of the house, T_C is the cold temperature and T_H is the hot temperature. Recall use absolutes temperatures(273.15+^\circ C). Replacing and solving for W:

W=Q_c\frac{T_H-T_C}{T_C}\\W=2000J\frac{312.15K-293.15K}{293.15K}\\W=129.63J

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3 years ago
In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and e
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Answer:

The  velocity is  v_t  =  0.02175 \  m/s

Explanation:

From the question we are told that

   The  mass of the bullet is  m_b  =  0.024 \  kg

    The initial speed of the bullet is  u_b  =  1200 \  m/s

   The mass of the target is  m_t  =  320 \  kg

    The  initial velocity of target is  u_t  =  0  \ m/s

    The  final velocity of the bullet is  is  v_b  =  910 \  m/s

   

Generally according to the law of momentum conservation we have that

      m_b *  u_b  +  m_t *  u_t  =  m_b *  v_b  +  m_t  *  v_t

=>   0.024  *  1200  +  320 *  0  =  0.024 *  910   +  320  *  v_t

=>    v_t  =  0.02175 \  m/s

3 0
3 years ago
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