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Firdavs [7]
3 years ago
15

Answer the following question below oki plzzz

Physics
1 answer:
klasskru [66]3 years ago
8 0
The answer is letter A.
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(,I hope this helps yoy)

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3 years ago
Problem:
kenny6666 [7]

Answer:

a) x = 1.5 *10⁻⁴cos(524πt) m

b) v = -1.5 *10⁻⁴(524π)sin(524πt) m/s

   a =  -1.5 *10⁻⁴(524π)²cos(524πt) m/s²

c) x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm

x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm

Explanation:

x = Acos(ωt)

ω = 2πf = 2π(262) = 524π rad/s

x = 1.5 *10⁻⁴cos(524πt)

v = y' = -Aωsin(ωt)

v = -1.5 *10⁻⁴(524π)sin(524πt)

a = v' = -Aω²cos(ωt)

a =  -1.5 *10⁻⁴(524π)²cos(524πt)

not sure about the last part as time is generally not given in mm

I will show at 1 second and at 0.001 s to try to cover bases

x(1) = 1.5 *10⁻⁴cos(524π(1))

x(1) = 1.5 *10⁻⁴cos(524π)

x(1) = 1.5 *10⁻⁴(1)

x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm

x(0.001) = 1.5 *10⁻⁴cos(524π(0.001))

x(0.001) = 1.5 *10⁻⁴cos(0.524π)

x(0.001) = 1.5 *10⁻⁴(-0.0753268)

x(0.001) = -1.129902...*10⁻⁵ m

x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm

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3 years ago
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Come in my free candy van located at Skellerupvej 43, 8600 Silkeborg, Denmark
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Answer:

Did you think you could pull that off-

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I live in Japan anyway so.. ウーフ-

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