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const2013 [10]
2 years ago
13

Need an explanation please. Brainliest will be awarded.

Physics
1 answer:
Maurinko [17]2 years ago
7 0

We're adding two vectors here.  The first is 300 Newtons to the right, which we can write as (300, 0), meaning 300 to the right, 0 up.

The second is 300 at let's say a 45 degree angle down.  For the components we have an isosceles right triangle with hypotenuse 300, so the components are both magnitude 300/√2 = 150√2.  So we can write this vector (150√2, -150√2), the negative sign because it points down in the y direction.

Adding is componentwise.  The resulting force is (300+150√2, -150√2).

That has square magnitude

r² = (300+150√2)² + (-150√2)² = 150² ( (2+√2)² + (√2)² )

= 150²( (6 + 4√2) + 2)

= 300²(2+√2)

so

r = 300 √(2+√2) Newtons

That's the answer; I'm not sure if your class expects a calculator approximation, which is 554.3 Newtons.

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Answer:

The resistance of the inductor at resonance is 258.76 ohms.

Explanation:

Given;

resistance of the resistor, R = 305 ohm

capacitance of the capacitor, C = 1.1 μF = 1.1 x 10⁻⁶ F

inductance of the inductor, L = 42 mH = 42 x 10⁻³ H = 0.042 H

At resonance the inductive reactance is equal to capacitive reactance.

\omega L = \frac{1}{\omega C}\\\\2\pi F_0 L =  \frac{1}{2\pi F_0 C}\\\\F_0 = \frac{1}{2\pi\sqrt{LC} }

Where;

F₀ is the resonance frequency

F_0 = \frac{1}{2\pi\sqrt{LC} } \\\\F_0 = \frac{1}{2\pi\sqrt{(0.024)(1.1*10^{-6})} }\\\\F_0 =980.4 \ Hz

The inductive reactance is given by;

X_l = 2\pi F_0 L\\\\X_l = 2\pi (980.4)(0.042) \\\\X_l = 258.76 \ ohms

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