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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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A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are
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Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

Coordinate of position of y=3t^2+2

Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

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The acceleration in x axis at 2 sec

a = 60i

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y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

We need to calculate the net force

F = ma

F = 3.00\times(60i+6j)

F=180i+18j

The magnitude of the force

|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

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