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Mandarinka [93]
3 years ago
12

A typical ceiling fan running at high speed has an airflow of about 2.00 ✕ 103 ft3/min, meaning that about 2.00 ✕ 103 cubic feet

of air move over the fan blades each minute.
Determine the fan's airflow in m3/s.
Physics
1 answer:
Leno4ka [110]3 years ago
8 0

Answer:

0.94 m³/s

Explanation:

From the question given above, the following data were obtained:

Air flow (in ft³/min) = 2×10³ ft³/min

Air flow (in m³/s) =.?

Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:

35.315 ft³/min = 1 m³/min

Therefore,

2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min

2×10³ ft³/min = 56.63 m³/min

Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:

1 m³/min = 1/60 m³/s

Therefore,

56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min

56.63 m³/min = 0.94 m³/s

Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.

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The key formula ===> Voltage = (current) x (resistance)

Plug in the numbers given ===> Voltage = (3.6 A) x (5.0 ohms)

Dooda multiplication ===> Voltage = 18 volts

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3 years ago
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A 2.2-m-long steel rod must not stretch more than 1.2 mm when it is subjected to a 8.5-kN tension force. Knowing that E = 200 GP
kodGreya [7K]

Answer:

a) 0.00996 m

b) 109090909 Pa

Explanation:

Unit conversions:

E = 200GPa = 200\times10^9 Pa

1.2 mm = 0.0012 m

8.5 kN = 8500 N

If the 2.2m rod cannot stretch more than 0.0012 m, its maximum strain is

\epsilon = \frac{\Delta L}{L} = \frac{0.0012}{2.2} = 0.000545455

With elastic modulus being E = 200 GPa, then its maximum stress must be

\sigma = E\epsilon = 200\times10^9*0.000545455 = 109090909 Pa

Knowing the tension force being F = 8500 N, we can calculate the appropriate cross section area

A = \frac{F}{\sigma} = \frac{8500}{109090909} = 7.79\times10^{-5}m^2

And its corresponding diameter is

A = \pi d^2/4

7.79\times10^{-5} = \pi d^2/4

d^2 = \frac{4*7.79\times10^{-5}}{\pi} = 9.92\times10^{-5}

d = \sqrt{9.92\times10^{-5}} = 0.00996 m \approx 1 cm

7 0
3 years ago
A rugby player passes the ball 7.88 m across the field, where it is caught at the same height as it left his hand. (Assume the p
damaskus [11]

Explanation:

Given

Range of ball =7.88 m

Initial speed=13.1 m/s

and we know

Range of Projectile is

R=\frac{u^2sin2\theta }{g}

13.1=\frac{13.1^2\times sin2\theta }{9.8}

7.88\times 9.8=13.1^2\times sin2\theta

sin2\theta =0.4499

2\theta =26.74^{\circ}

\theta =13.37 ^{\circ}

(b)other angle would be complementary of \theta

=90-\theta =76.63 ^{\circ}

8 0
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How is periodicity of properties dependent upon number of protons in an atom?
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Depending on the charge.
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4 years ago
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Suppose there was a star with a parallax angle of 1 arcsecond. How far away would it be? Select all that apply.
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Answer:

option E

Explanation:

given,                          

Parallax angle(d) = 1 arcsecond

using Parallax formula                  

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 p is the parsecs angle which is measured in 1 arcsecond

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now,                                            

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    1 parsec = 3.26 light year

hence, the answer will be option E

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