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3 years ago

The SI unit of average spped and velocity.

1 answer:

3 0

meter per second is the main answer of both

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How our fish adapted to thier enviorment?

lubasha [3.4K]

They use their gills to swim around and they sleep with their eyes open

6 0

4 years ago

Charge is distributed uniformly throughout the volume of an infinitely long solid Cylinder of radius R what is the electric fiel

Arada [10]

Solution :

Let us consider the Gaussian surface that is in the form of a cylinder having a radius of r and a length of A which is $\text{coaxial with the charged cylinder}$ .

The charged enclosed by the cylinder is given by,

$q=\rho V$ (here, V = $\pi r^2l$ is the volume of the cylinder)

$=\pi r^2lp$

If $\rho$ is positive, then the electric field lines moves in the radial outward direction and is normal to Gaussian surface which is distributed uniformly.

Therefore, total flux through Gaussian cylinder is :

$\phi=EA_{cyl}$

$=E(2\pi rl)$

Now using Gauss' law, we get

$2\pi \epsilon_0rlE = \pi r^2lp$

or $E=\frac{\rho r}{2 \epsilon_0}$

Therefore, the electric field is $E=\frac{\rho r}{2 \epsilon_0}$

Hence, option (d) is correct.

8 0

3 years ago

A 330-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,110 A. If the conductor is copper

vodka [1.7K]

Answer:

t = 402 years

Explanation:

To find the number of year that electrons take in crossing the complete transmission line, you first calculate the drift speed of the electrons. Then, you use the following formula for the current in a wire:

$I=nqv_dA$ (1)

n: number of mobile charge carrier per volume = 8.50*10^28 e/m^3

q: charge of the electron = 1.6*10^-19 C

vd: drift velocity of electron in the metal = ?

A: cross sectional area of the wire = π r^2 = π (0.02m/2)^2 = 3.1415*10^-4 m^2

I: current in the wire = 1110 A

You solve the equation (1) for vd:

$v_d=\frac{I}{nqA}=\frac{110A}{(8.50*10^{28}m^{-3})(1.6*10^{-19}C)(3.1415*10^{-4}m^2)}\\\\v_d=2.59*10^{-4}m/s$

Next, you calculate the time by using the information about the length of the line transmission:

$x=v_dt\\\\x=330km=330000m\\\\t=\frac{x}{v_d}=\frac{330000m}{2.59*10^{-4}m/s}=1,270,184,865s\\\\1,270,184,865s*\frac{1\ year}{3,156,107}=402.45\ years$

hence, the electrons will take aproximately 402 years in crossing the line of transmission

6 0

3 years ago

A television channel is assigned the frequency range from 54 MHz to 60 MHz. A series RLC tuning circuit in a TV receiver resonat

Stella [2.4K]

Answer:

A.) L = 0.37 μH B) 7.61 Ω

Explanation:

A) At resonance, the circuit behaves like it were purely resistive , so the reactance value must be 0.

So, the following condition must be met:

ω₀*L = 1/ (ω₀*C) ⇒ ω₀² = 1/LC

We know that, for a sinusoidal source, there exists a fixed relationship between the angular frequency ω₀ and the frequency f₀, as follows:

ω₀ = 2*π*f₀

⇒ (2*π*f₀)² = 1/(L*C)

Replacing by the givens (f₀, C), we can solve for L:

L = 1 /((2*π*f₀)²*C) = 1/(2*π*57*10⁶)² Hz²*21*10⁻¹² f = 0.37 μH

b) At resonance, the current can be expressed as follows:

I₀ = V/Z = V/R

We need to find the minimum value of R that satisfies the following equation:

I = 0.5 I₀ = 0.5 V/R = V/Z

⇒ 0.5/R = 1/√(R²+X²)

Squaring both sides , we have:

(0.5)²/R² = 1/ (R²+X²)

⇒ 0.25 (R²+X²) = R² ⇒ R² = X² / 3

We need to find the value of R that satisfies the requested condition througout the frequency range.

So, we need to find out the value of the reactance X in the lowest and highest frequency, as follows:

Xlow = ωlow * L - 1/(ωlow*C)

⇒ Xlow = ( (2*π*54*10⁶)*0.37*10⁻⁶) - 1/ ((2*π*54*10⁶)*21*10⁻¹²) = -14.81Ω

Xhi = ωhi * L - 1/(ωhi*C)

⇒ Xhi = ( (2*π*60*10⁶)*0.37*10⁻⁶) - 1/ ((2*π*60*10⁶)*21*10⁻¹²) = 13.18Ω

For these reactance values, we can find the corresponding values of R as follows:

Rlow² = Xlow²/3 = (-14.81)²/3 = 75Ω² ⇒ Rlow = 8.55 Ω

Rhi² = Xhi² / 3 = (13.18)²/3 = 56.33Ω² ⇒Rhi = 7.61 Ω

The minimum value of R that satisfies the requested condition is R= 7.61ΩΩ.

3 0

3 years ago

A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 280 Ω at room temperature, what is

monitta

Answer:

r = 17.05 cm

Explanation:

Given that,

Length of silicon bar is 4 cm or 0.04 m

Resistance of the bar is 280 ohms

We know that the resistivity of the silicon is 640 Ωm

We need to find the cross-sectional radius of the bar. Let it is r.

Using definition of resistance of an object. It is given by :

$R=\rho\dfrac{l}{A}$

A is area of bar, A = πr²

So,

$R=\rho\dfrac{l}{\pi r^2}\\\\r^2=\dfrac{\rho l}{R\pi}\\\\r^2=\dfrac{640\times 0.04}{280\pi}\\\\r=0.1705\ m\\\\r=17.05\ cm$

So, the cross-sectional radius of the bar is 17.05 cm.

4 0

4 years ago

The SI unit of average spped and velocity.​

The SI unit of average spped and velocity.