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3 years ago

Express 3 x 10-9s in milliseconds

Physics

2 answers:

Sveta_85 [38]3 years ago

7 0

1 second = 10^3 milliseconds

Multiply the value 3x10^-9 by 10^3 which would give you 3x10^-6

Art [367]3 years ago

3 0

Answer:

3 × 10 to the power minus 6 milliseconds

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The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?

Yuki888 [10]

Answer:

-.457 m/s^2

Explanation:

Actual weight = 60 .3 (9.81) = 591.54 N

Accel of lift changes this to 60.3 ( 9.81 - L) where L - accel of lift

60.3 ( 9.81 - L ) = 564

solve for L = .457 m/s^2 DOWNWARD

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2 years ago

If two objects are the same size, will they have the same mass?

jeka57 [31]

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3 years ago

You are trying to overhear a most interesting conversation, but from your distance of 10.0 m , it sounds like only an average wh

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Answer:

r₂=0.1 m

Explanation:

Given that

r₁= 10 m , β₁ = 20 dB

At r₂ ,β₂= 60 dB

As we know that intensity level of sound given as

$\beta =10\ log\dfrac{I}{10^{-12}}$

$\beta _1=10\ log\dfrac{I_1}{10^{-12}}$

$20=10\ log\dfrac{I_1}{10^{-12}}$

10² x 10⁻¹² = I₁

I₁=10⁻¹⁰ W/m²

$\beta _2=10\ log\dfrac{I_2}{10^{-12}}$

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10⁶ x 10⁻¹² = I₂

I₂ = 10⁻⁶ W/m²

I₁=10⁻¹⁰ W/m²

P = I A

P=Power ,I =Intensity ,A=Area

$\dfrac{I_1}{I_2}=\dfrac{r^2_2}{r^2_1}$

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3 years ago

Everyone's favorite flying sport disk can be approximated as the combination of a thin outer hoop and a uniform disk, both of di

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Answer:

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md = 0.05 kg

mh = 0.12 kg

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The length of the boomerang is:

$L_{b} =\sqrt{\frac{12I}{m} } =\sqrt{\frac{12*2.7x10^{-3} }{0.245} } =0.364m$

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3 years ago

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3 years ago