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Fynjy0 [20]
3 years ago
5

The inductor in the RLC tuning circuit of an AM radio has a value of 250 mHmH . You may want to review (Pages 857 - 860) . Part

A What should be the value of the variable capacitor in the circuit to tune the radio to 731 kHz kHz
Physics
1 answer:
ANEK [815]3 years ago
6 0

Answer:

The value of variable capacitor is 1.89 \times 10^{-13} F

Explanation:

Given :

Inductance L = 250 \times 10^{-3} H

Frequency f = 731 \times 10^{3} Hz

According to the cutoff frequency,

   f = \frac{1}{2\pi \sqrt{LC} }

Now we find the value of capacitance,

  C = \frac{1}{4\pi ^{2} f^{2}  L }

  C = \frac{1}{4\times 9.85 \times (731 \times 10^{3} )^{2} \times 250 \times 10^{-3}  }

  C = 1.89 \times 10^{-13} F

Therefore, the value of variable capacitor is 1.89 \times 10^{-13} F

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A satellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution abou
kupik [55]

Answer:

The acceleration of the satellite is 0.87 m/s^{2}

Explanation:

The acceleration in a circular motion is defined as:

a = \frac{v^{2}}{r}  (1)

Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

v = \frac{2 \pi r}{T} (2)

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite (1.50x10^{7} m) and the Earth radius (6.38x10^{6} m) :

r = 1.50x10^{7} m+6.38x10^{6}m

r = 21.38x10^{6}m

Then, equation 2 can be used:

T = 8.65 hrs \cdot \frac{3600 s}{1hrs} ⇒ 31140 s

v = \frac{2 \pi (21.38x10^{6}m)}{31140s}

v = 4313 m/s

Finally equation 1 can be used:

a = \frac{(4313m/s)^{2}}{21.38x10^{6}m}    

a = 0.87 m/s^{2}

Hence, the acceleration of the satellite is 0.87 m/s^{2}

6 0
3 years ago
How can meteorologists use the jet stream to predict the weather ?
barxatty [35]
<span>Jet streams act as an invisible director of the atmosphere and are largely responsible for changes in the weather across the globe.
Hope this helps</span>
5 0
3 years ago
Name three bright Saturn ring features, and explain why they are so bright.
Leviafan [203]

The F Ring, the Cassini Division, and the C Ring are bright ring features. They are bright due to the low concentration of materials within them, which allows sunlight to shine through.


6 0
3 years ago
Which of these will always produce a magnetic field?
vovikov84 [41]

Answer:

Technically everything has somewhat of a magnetic field. I guess

7 0
3 years ago
One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren
katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

\sigma = \frac{F}{A}

Where,

\sigma =Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

\sigma_y = \frac{F_y}{A}

where,

\sigma_y =Shear strength

F_y = Shear Force

A_0 =Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}

Resolving for F,

F= 1399.5N

PART B) We need here to apply the shear strength equation, then

\sigma_y = \frac{F_y}{A}

210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}

F_y = 945N

In such a way that the material is more resistant to tensile strength than shear force.

6 0
3 years ago
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