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3 years ago

Complete this equation that represents the process of nuclear fusion.

Physics

1 answer:

Mars2501 [29]3 years ago

8 0

You need to have the exact amount of atomic number and mass number on each side of the arrow. on the left of the equation Ra has 226 mass and 88 atomic. on the right the He with the atomic number of 2 and the mass number of 4. you’d minus the atomic number on the left with the atomic number on the right (88-2) getting 86 and minus the mass number on the left with the mass number on the right (226-2) which is 224

this means that A = 224 and B = 86

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2 years ago

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2 years ago

In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s

sladkih [1.3K]

Answer:

quality factor (Q) = 69.99

inductor = 1.591 x 10⁻⁴ H

capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R = 10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW) $= \frac{R}{2\pi L}$

$BW = \frac{R}{2\pi L}$

make L (inductor) the subject of the formula

$L = \frac{R}{2\pi *BW} = \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH$

$F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}$

make C (capacitor) the subject of the formula

$C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F$

quality factor (Q) $= \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99$

quality factor (Q) = 69.99

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3 years ago

Consider a mechanical clutch that consists of two heavy disks that can engage or disengage. At the beginning disk 1 with mass m1

miv72 [106K]

Answer:

w = 4,786 rad / s , f = 0.76176 Hz

Explanation:

For this problem let's use the concept of angular momentum

L = I w

The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved

Initial Before sticking

L₀ = 0 + I₂ w₂

Final after coupling

$L_{f}$ = (I₁ + I₂) w

The moments of inertia of a disk with an axis of rotation in its center are

I = ½ M R²

How the moment is preserved

L₀ = $L_{f}$

I₂ w₂ = (I₁ + I₂) w

w = w₂ I₂ / (I₁ + I₂)

Let's reduce the units to the SI System

d₁ = 60 cm = 0.60 m

d₂ = 40 cm = 0.40 m

f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz

Angular velocity and frequency are related.

w₂ = 2 π f₂

w₂ = 2π 3.33

w₂ = 20.94 rad / s

Let's replace

w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)

w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)

Let's calculate

w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)

w = 20.94 1.28 / 5.6

w = 4,786 rad / s

Angular velocity and frequency are related.

w = 2π f

f = w / 2π

f = 4.786 / 2π

f = 0.76176 Hz

7 0

3 years ago

Complete this equation that represents the process of nuclear fusion.​

Complete this equation that represents the process of nuclear fusion.