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Gnesinka [82]
2 years ago
12

The density of a block of wood is 2.7 g/cm and a mass of 583.3 g.

Physics
1 answer:
Norma-Jean [14]2 years ago
3 0

Answer:

C. 216.0 cm³

Explanation:

  • formula to solve this:

                                          Volume = mass ÷ density

                                          Volume = 583.3 g ÷ 2.7 g/cm

                                          Volume = 216 cm³

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2 years ago
Determine the energy in joules of a photon whose frequency is 3.55 x10^17 hz
asambeis [7]
By using the Plancks-Einstein equation, we can find the energy;
E = hf
where h is the plancks constant = 6.63 x 10⁻³⁴
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3 years ago
VERY EASY QUESTION!!!!!! 20 POINTS!!!!!!!!
slamgirl [31]
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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

5 0
3 years ago
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