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3 years ago

What is the critical angle - easy definition

1 answer:

8 0

the angle of incidence beyond which rays of light passing through a denser medium to the surface of a less dense medium are no longer refracted but totally reflected.

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A straight wire of length 0.53 m carries a conventional current of 0.2 amperes. What is the magnitude of the magnetic field made

olga55 [171]

Explanation:

It is given that,

Length of wire, l = 0.53 m

Current, I = 0.2 A

(1.) Approximate formula:

We need to find the magnitude of the magnetic field made by the current at a location 2.0 cm from the wire, r = 2 cm = 0.02 m

The formula for magnetic field at some distance from the wire is given by :

$B=\dfrac{\mu_oI}{2\pi r}$

$B=\dfrac{4\pi \times 10^{-7}\times 0.2\ A}{2\pi \times 0.02\ m}$

B = 0.000002 T

$B=10^{-5}\ T$

(2) Exact formula:

$B=\dfrac{\mu_oI}{2\pi r}\dfrac{l}{\sqrt{l^2+4r^2} }$

$B=\dfrac{\mu_o\times 0.2\ A}{2\pi \times 0.02\ m}\times \dfrac{0.53\ m}{\sqrt{(0.53\ m)^2+4(0.02\ m)^2} }$

B = 0.00000199 T

B = 0.000002 T

Hence, this is the required solution.

4 0

3 years ago

30 Pts! Answer fast please! mass= 1 kg KE= 18 Joules What is the velocity?

makvit [3.9K]

Its 21!! sorry i was late! :)

5 0

3 years ago

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

A 1.50-m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance x fro

MArishka [77]

Answer:

Resistance = 3.35* $10^{-4}$ Ω

Explanation:

Since resistance R = ρ $\frac{L}{A}$

whereas $\rho(x) = a + bx^2$

resistivity is given for two ends. At the left end resistivity is $2.25* 10^{-8}$ whereas x at the left end will be 0 as distance is zero. Thus

$2.25*10^{-8} = a + b(0)^2\\ 2.25*10^{-8} = a + 0 \\2.25*10^{-8} = a$

At the right end x will be equal to the length of the rod, so $x = 1.50\\8.50*10^{-8} = (2.25*10^{-8}) + ( b* (1.50)^2 )\\8.50*10^{-8} - (2.25*10^{-8}) = b*2.25\\\frac{6.25*10^{-8}}{2.25} = b\\b = 2.77 *10^{-8}$