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Hunter-Best
3 months ago
6

A ceiling fan has five blades, each with a mass of 0.34 kg and a length of 0.66 m. The fan is operating in its "low" setting at

which the angular speed is 9.4 rad/s. If the blades can be approximated as uniform thin rods that rotate about one end, what is the total rotational kinetic energy of the five blades?

Physics
1 answer:
Yuliya22 [3.3K]3 months ago
8 0
The cumulative rotational kinetic energy of the five blades amounts to 10.9J. Please refer to the attached documents for further information.
You might be interested in
A 1.97-pF capacitor with a plate area of 5.86 cm2 and separation between the plates of 2.63 mm is connected to a 9.0-V battery a
inna [3103]

If the plate separation is modified after the battery is disconnected, the updated distance between plates is 9.21 mm

If changes are made while the battery remains connected, the new separation becomes 0.11 mm

The capacitance for an air-filled parallel plate capacitor can be expressed as:

C=\frac{\epsilon_0A}{d}

In this equation, \epsilon_0 refers to the permittivity of free space, A stands for the plate area, and D represents the separation distance.

Thus,

C \alpha \frac{1}{d}.......(1)

Therefore, should the distance between the plates shift from d₁ to d₂, the capacitance ratio in both scenarios can be represented as:

\frac{C_1}{C_2} =\frac{d_2}{d_1}......(2)

Scenario (i)

When the capacitor is fully charged and then disconnected from the battery before adjusting the plate distance, the charge will remain steady while the capacitance varies.

The initial energy E₁ stored in the capacitor can be expressed as:

E_1=\frac{Q^2}{2C_1}......(3)

Once the separation changes to d₂, capacitance becomes C₂, but the charge Q remains unchanged.

Thus,

E_2=\frac{Q^2}{2C_2}......(4)

By dividing equation (4) by (3),

\frac{E_2}{E_1} =\frac{C_1}{C_2}

According to equation (2),

\frac{E_2}{E_1} =\frac{C_1}{C_2}=\frac{d_2}{d_1}

This results in a 3.5 fold increase in energy.

\frac{E_2}{E_1} =\frac{d_2}{d_1}=3.5\\ d_2=3.5*2.63 mm\\ =9.205 mm=9.21 mm

Scenario (2)

If the capacitor is kept connected to the power source, the voltage V across the plates will remain unchanged.

The initial energy is described as

E_1=\frac{1}{2} C_1V^2......(5)

The final energy when the plate separation transitions to d₂ can be written as:

E_2=\frac{1}{2} C_2V^2.....(6)

Referencing equations (5) and (6)

\frac{E_2}{E_1} =\frac{C_2}{C_1}

From equation (2),

\frac{E_2}{E_1} =\frac{C_2}{C_1}=\frac{d_1}{d_2}

Thus, in this particular scenario,

\frac{E_2}{E_1} =\frac{d_1}{d_2}\\d_2=\frac{d_1}{3.5} \\ =\frac{2.63 mm}{3.5} \\ =0.109 mm=0.11 mm

Therefore,

Adjusting plate separation after battery disconnection yields 9.21 mm

If modified while connected, the new separation measures 0.11 mm





6 0
3 months ago
A plane takes off in san francisco at noon and flies toward the southeast. an hour later, it is 400 kilometers east and 300 kilo
Keith_Richards [3271]
This initial stage represents a right triangle. If the aircraft is positioned 400 km to the east and 300 km to the south of the origin while traveling in a straight path, you can form a right triangle with sides measuring 300, 400, and c. You might notice that these dimensions correspond to multiples of the Pythagorean triple 3, 4, 5, meaning that the length of c is 500 km. Alternatively, you would indicate
c^2 = 300^2+400^2 \\ c = 500.

For the second step, assuming I am correctly understanding "degrees south of east," it involves calculating the angle between the horizontal line indicating east and the trajectory of the aircraft. I created a diagram illustrating this (see attached). You could employ a trigonometric function related to one of the angles to find the solution. I selected
tan(C)= \frac{3}{4} \\ arctan( \frac{3}{4})=36.870. Therefore, I deduce that the angle is 37° south of east.

5 0
2 months ago
A small 175-g ball on the end of a light string is revolving uniformly on a frictionless surface in a horizontal circle of diame
kicyunya [3294]
For motion in a circle.

Centripetal acceleration is calculated as mv²/r = mω²r

where v represents linear velocity, r equals radius which is diameter/2 equating to 1/2 or 0.5m

. Here, m is the mass of the object, which is 175g or 0.175kg.

The angular speed, ω, is derived from Angle covered / time

                         = 2 revolutions per 1 second

                         = 2 * 2π  radians for each second

                         = 4π  radians per second

Thus, Centripetal Acceleration = mω²r = 0.175*(4π)² * 0.5. Utilize a calculator

                                                         ≈13.817  m/s²

. The acceleration's magnitude is approximately 13.817  m/s² and it is oriented towards the center of the circular path.

The tension in the string equates to m*a

                                   = 0.175*13.817

                                   = 2.418 N
5 0
3 months ago
Read 2 more answers
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