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Hitman42 [59]
2 years ago
6

A runner is at rest at the starting line of a race. She accelerates at 3.3m/s/s. What is her position after 7.5s? Round your ans

wer to the nearest tenth of a meter.
Physics
1 answer:
natulia [17]2 years ago
3 0

Answer:

sorry ‍♂️I don't know the answer

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Which of the following objects has the greatest density
maks197457 [2]
The density is determined on the steepness of the slope. The greater the density is bases upon the steepest slope. To conclude, I'd say Line A has the steepest slope therefore has the greatest density.
6 0
3 years ago
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A series RLC circuit has a resonant frequency = 6.00 kHz. When it is driven at a frequency = 8.00 kHz, it has an
ANEK [815]

The resistance (R) of the circuit is 707.1 ohms and the inductance (L) is 0.032 H.

<h3>Resistance of the circuit</h3>

For the phase constant of 45⁰, impedance is equal to the resistance of the circuit.

Z= R\sqrt{2} \\\\R  = \frac{Z}{\sqrt{2} } \\\\R = \frac{1000}{\sqrt{2} } = 707.1 \ ohms

<h3>Resonant frequency</h3>

f = \frac{1}{2\pi \sqrt{LC} } \\\\6000 = \frac{1}{2\pi \sqrt{LC} } \\\\2\pi(6000) = \frac{1}{\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi (6000)} \\\\LC = (\frac{1}{2\pi (6000)})^2\\\\LC = 7.034 \times 10^{-10} \\\\ C = \frac{7.034 \times 10^{-10} }{L} ---(1)

<h3>At driven frequency</h3>

X_l- X_c = R\\\\\omega L - \frac{1}{\omega C}  = 707.1\\\\2\pi f L -  \frac{1}{2\pi fC} = 707.1\\\\2\pi (8000) L - \frac{1}{2\pi (8000) C } = 707.1\ \ --(2)\\\\

<em>solve 1 and 2 together</em>

2\pi(8000) L - \frac{L}{2\pi (8000)(7.034 \times 10^{-10})} = 707.1\\\\50272L - 28279.48L = 707.1\\\\L = 0.032 \ H

Learn more about impedance of RLC circuit here: brainly.com/question/372577

7 0
2 years ago
Please help me. I'll mark brainiest
valkas [14]
1-d 2-a. This is not my field of expertise but this is my assumption
5 0
2 years ago
You're using a monochromatic beam of light with wavelength 500 nm in an interferometer. What is the minimum distance you would n
xxTIMURxx [149]

Answer:

The minimum distance moved by mirror is 125 nm

Explanation:

Given:

Wavelength of light \lambda = 500 \times 10^{-9} m

Here given in question, spot of constructive interference change to a destructive interference.

Hence minimum distance moved by one of mirror is given by,

   t = \frac{1}{2}(\frac{\lambda}{2} )

Where t = distance moved by mirror

   t = \frac{\lambda }{4}

   t = \frac{500\times 10^{-9} }{4}

   t = 125 \times 10 ^{-9}

Therefore, the minimum distance moved by mirror is 125 nm

4 0
3 years ago
a 150 kg object takes 1.5 minutes to travel a 2,500 meter straight path. It begins the trip traveling 120 m/s and decelerates to
Firlakuza [10]

Answer:

-1.11 m/s²

Explanation:

v = at + v₀

Given v = 20 m/s, v₀ = 120 m/s, and t = 1.5 min = 90 s:

20 m/s = a (90 s) + 120 m/s

a = -1.11 m/s²

7 0
2 years ago
Read 2 more answers
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