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yawa3891 [41]
3 years ago
7

1. Which type of wire would be a better conductor of an electrical current?

Physics
1 answer:
Lelechka [254]3 years ago
7 0

Answer:

1. a

2. b

3. b

Explanation:

1.

Resistance is the property of a conductor to offer resistance to the flow of current. The lower the resistance better is  the conductivity of wire.

We know that the resistance of a wire depends on several factor which are inter-connected by an equation as:

R=\rho.\frac{l}{a}

where:

R = resistance of the wire

l = length of the wire

a= cross sectional area of the wire  

from the above relation we observe that

R\propto l

R\propto \frac{1}{a}

  • Also when the temperature of the wire is significantly high then the lattice vibration cause obstruction in the path of the flowing charges and reduce the current flow.

2.

As the collision between the electrons and protons increases the speed of the flow of charges will decrease because the opposite charges attract each other and as we know that electrical current is the rate of flow of charge.

3.

Heating up of wire due to sunlight will cause lattice vibration in the conductor and will obstruct the movement of the charges which build up electric current, hence increasing the resistance of conductivity.

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Answer:

The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

Explanation:

Given that,

First charge q_{1}= 2.9\mu C

Second chargeq_{2}= 2.9\mu C

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

E_{1}=\dfrac{kq}{r'^2}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}

E_{1}=52200=52.2\times10^{3}\ N/C

For E₂,

Using formula of electric field

E_{1}=\dfrac{kq}{r^2}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}

E_{2}=104400=104.4\times10^{3}\ N/C

We need to calculate the horizontal electric field

E_{x}=E_{1}\cos\theta

E_{x}=52.2\times10^{3}\times\cos45

E_{x}=36910.97=36.9\times10^{3}\ N/C

We need to calculate the vertical electric field

E_{y}=E_{2}+E_{1}\sin\theta

E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45

E_{y}=141310.97=141.3\times10^{3}\ N/C

We need to calculate the net electric field

E_{net}=\sqrt{E_{x}^2+E_{y}^2}

Put the value into the formula

E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}

E_{net}=146038.69\ N/C

E_{net}=146.03\times10^{3}\ N/C

We need to calculate the direction of electric field

Using formula of direction

\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}

\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})

\theta=75.36^{\circ}

Hence, The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

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WHAT IS THE DENSTY OF ALL THE LAYER IN THE ATMOSPHERE
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Answer:

The troposphere starts at the Earth's surface and extends 8 to 14.5 kilometers high (5 to 9 miles). This part of the atmosphere is the most dense. Almost all weather is in this region.

Explanation:

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Answer:

A -TRUE

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3) A dock worker pushes a 72 kg crate up a 2.0 m high,
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Work done on the crate is 1411.2 J

Explanation:

Work done is defined as the product of force and the distance moved by the object. The unit of work done is in joules and denoted by the symbol J.

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where F represents the force and d represents the distance moved by the object.

mass = 72 kg , distance moved by the object is given by 2.0 m

Force F = mass * gravity = 72 * 9.8

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P = (2*19) + (2*3)

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