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seropon [69]
3 years ago
9

At a sudden contraction in a pipe the diameter changes fromD1toD2. The pressure drop,Δp, which develops across the contraction i

s a function ofD1andD2, as well as the velocity,V, in the larger pipe, and the fluid density, rho, and viscosity,μ. UseD1,V, andμas repeating variables to determine a suitable set of dimensionless parameters. Why would it be incorrect to include the velocity in the smaller pipe as an additional variable?

Physics
1 answer:
Sever21 [200]3 years ago
5 0

Answer: The working and answer can be viewed from the screenshots below. Thanks

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find the period of a simple pendulum of 1m length placed on earth and on moon g on moon =1.67m/s² g on earth=10m/s²
Ierofanga [76]

Answer:

T_{m } = 4.86 s

T_{e} = 1.98 s

Explanation:

<u><em>Given:</em></u>

Length = l = 1 m

Acceleration due to gravity of moon = g_{m} = 1.67 m/s²

Acceleration due to gravity of Earth = g_{e} = 10 m/s²

<u><em>Required:</em></u>

Time period = T = ?

<u><em>Formula:</em></u>

T = 2π \sqrt{\frac{l}{g} }

<u><em>Solution:</em></u>

<u>For moon</u>

<em>Putting the givens,</em>

T = 2(3.14) \sqrt{\frac{1}{1.67} }

T = 6.3 \sqrt{0.6}

T = 6.3 × 0.77

T = 4.86 sec

<u>For Earth,</u>

<em>Putting the givens</em>

T = 2π \sqrt{\frac{1}{10} }

T = 2(3.14) \sqrt{0.1}

T = 6.3 × 0.32

T = 1.98 sec

3 0
3 years ago
In an atomic clock there are approximately 9.193 × 109oscillations of the specified light emitted by cesium-133 atoms. The text
Aleks [24]

Explanation:

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In 1.892 x 10¹¹ second ,  change in oscillation is 9.193 × 10⁹ oscillation

in one second change in oscillation = (9.193 / 1.892 ) x 10⁹⁻¹¹

=  4.859 x 10⁻² oscillations .

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podryga [215]

Answer:

A

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On a velocity_time graph

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Ksenya-84 [330]

Answer:is this a question??? I’m so confused

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Answer:

i  d  k

Explanation:

7 0
3 years ago
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