He is best known for publishing an almanac based on my astronomical calculations. Who is it?

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 thread holds two carts together on a frictionless surface. A spring is compressed

Goryan [66]

Answer:

Velocity on the right side of the cart $=0.09\ ms^{-1}$

Explanation:

Given

⇒The mass on the left of the cart $m_1=1.5\ kg$

Its velocity $v_1=27\ cm/s$ , $v_1=\frac{27}{100}=0.27\ m/s$

⇒Mass on the right of the cart $m_2=4.5\ kg$

Velocity $=?$ We have to find $v_2$

From

The law of conservation of linear momentum:

We can say that.

Initial momentum will equalize the final momentum.

And momentum is the product of mass and its velocity.

Assigning one of its velocity as negative because both are in different direction.

Lets call $v_1=-0.27m/s$

Recalling the formula and plugging the values.

$m_1(-v_1)+m_2v_2=0$

$v_2=-\frac{m_1(-v_1)}{m_2} =-\frac{1.5\times -0.27}{4.5} =0.09\ m/s$

So the velocity of the cart on the right side that has a mass of $4.5\ kg$ is $0.09\ ms^{-1}$

4 0

4 years ago

If a 20kg mass hangs from a spring, whose elastic constant is 1800 N / m, the value of the spring elongation is

almond37 [142]

Explanation:

F = kx

mg = kx

(20 kg) (10 m/s²) = (1800 N/m) x

x = 0.11 m

4 0

3 years ago

An aluminum wing on a passenger jet is 30 m long when its temperature is 27 C. At what temperature would the wing be 0.03 shorte

andrezito [222]

Answer:2000

Explanation:

6 0

3 years ago

A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal ener

Elina [12.6K]

Answer:

Explanation:

From the given information:

The initial PE $(PE)_i$ = m×g×h

= 5 kg × 9.81 m/s² × 10 m

= 490.5 J

The change in Potential energy P.E of the box is:

ΔP.E = $P.E_f -P.E_i$

ΔP.E = 0 - $(PE)_i$

ΔP.E = $-P.E_i$

If we take a look at conservation of total energy for determining the change in the internal energy of the box;

$\Delta P.E + \Delta K.E + \Delta U = 0$

$\Delta U = -\Delta P.E - \Delta K.E$

this can be re-written as:

$\Delta U =- (-\Delta P.E_i) - \Delta K.E$

Here, K.E = 0

Also, 70% goes into raising the internal energy for the box;

Thus,

$\Delta U =(70\%) \Delta P.E_i-0$

$\Delta U =(0.70) (490.5)$

ΔU = 343.35 J

Thus, the magnitude of the increase is = 343.35 J

7 0

3 years ago