Please help!!!! due today!!

Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter

alukav5142 [94]

Answer: 0.258

Explanation:

The resistance $R$ of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. For aluminium is $\rho_{Al}=2.65(10)^{-8}m\Omega$ and for copper is $\rho_{Cu}=1.68(10)^{-8}m\Omega$

$l$ is the length of the wire, which in the case of aluminium is $l_{Al}=12m$ , and in the case of copper is $l_{Cu}=30m$

$s$ is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

$s=\pi{(\frac{d}{2})}^{2}$ (2) Where $d$ is the diameter of the circumference.

For aluminium wire the diameter is $d_{Al}=2.5mm=0.0025m$ and for copper is $d_{Cu}=1.6mm=0.0016m$

So, in this problem we have two transversal areas:

For aluminium:

$s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}$

$s_{Al}=0.000004908m^{2}$ (3)

For copper:

$s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}$

$s_{Cu}=0.00000201m^{2}$ (4)

Now we have to calculate the resistance for each wire:

Aluminium wire:

$R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}$ (5)

$R_{Al}=0.0647\Omega$ (6) Resistance of aluminium wire

Copper wire:

$R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}$ (6)

$R_{Cu}=0.250\Omega$ (7) Resistance of copper wire

At this point we are able to calculate the ratio of the resistance of both wires:

$Ratio=\frac{R_{Al}}{R_{Cu}}$ (8)

$\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}$ (9)

Finally:

$\frac{R_{Al}}{R_{Cu}}=0.258$ This is the ratio

3 0

3 years ago

The magnetic field at the earth's surface can vary in response to solar activity. During one intense solar storm, the vertical c

Flauer [41]

Answer:

$EMF = 33880 Volts$

Explanation:

As per Faraday's law of Electromagnetic induction we know that

Rate of change in magnetic flux will induce EMF in the closed conducting loop

so we have

$\phi = B.A$

now we have

$A = (110 \times 10^3)(110 \times 10^3)$

$A = 1.21 \times 10^{10}$

now we have

$\phi = B(1.21 \times 10^{10})$

now the induced EMF through this loop is given as

$EMF = (\frac{dB}{dt})(1.21 \times 10^{10})$

$EMF = (2.8 \times 10^{-6})(1.21 \times 10^{10})$

$EMF = 33880 Volts$

5 0

3 years ago

8 points

Lunna [17]

The pilot might be correct (I think), because, if the gravity of the planet is strong, then the planet’s gravity will pull the spaceship into its orbit, so the engines don’t need to be on for the ship to get pushed toward the planet.

4 0

3 years ago

Read 2 more answers

It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$