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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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I wanna say its A . I could be wrong but im almost 100 percent sure that its A wood
4 0
3 years ago
A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is
lys-0071 [83]

Answer:

\omega_f=571.42\ rpm

Explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string, a=1.5\ m/s^2

Displacement, d = 1.3 m

The angular acceleration is given by :

\alpha =\dfrac{a}{r}

\alpha =\dfrac{1.5}{0.033}

\alpha =45.46\ rad/s^2

The angular displacement is given by :

\theta=\dfrac{d}{r}

\theta=\dfrac{1.3}{0.033}

\theta=39.39\ rad

Using the third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta

Here, \omega_i=0

\omega_f=\sqrt{2\alpha \theta}

\omega_f=\sqrt{2\times 45.46\times 39.39}

\omega_f=59.84\ rad/s

Since, 1 rad/s = 9.54 rpm

So,

\omega_f=571.42\ rpm

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.

5 0
3 years ago
1. In Newton’s ring experiment, the diameter of the 5th ring is 0.30 cm and diameter of 15th the ring is 0.62 cm. Find the diame
IgorC [24]

Answer:

Diameter of Newton’s 5th ring = 0.30 cm

Diameter of Newton’s 15th ring = 0.62 cm

Diameter of Newton’s 25th ring = ?

From Newton’s rings experiment we infer that

D2n+m − D2n = 4λmR

For the 5th and 15th rings we have

D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)

For 15th and 25th rings

D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)

We equate the two derivatives

Equation (2) = Equation (1)

D225 − D215 = D215 − D25

D225 = 2D215 – D25

Substituting the values into the equation

D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2

D25 = 0.8239 cm

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What is the difference between balanced force and action reaction force
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Answer:

Balanced forces are equal and opposite forces that act on the same object. ... Action-reaction forces are equal and opposite forces that act on different objects, so they don't cancel out. In fact, they often result in motion.

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