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

Answer: B

Explanation: I choose B because when you pull back something, the force it exerts when released is greater than when you hold it back, especially for a rubber band.

Answer:

The force will rise in direct relation to the mass of the objects

Explanation:

The gravitational acceleration remains constant. It is measured in meters per second squared or m/s². The average value is 9.81 m/s², calculated from observations made on varying surfaces. In reality, the acceleration can vary based on the geographical shape of the Earth relative to the earth's magnetic field and gravitational force.

For instance, if a single washer weighs 20 kg, with the gravity at 9.81 m/s², the weight would be:

F = ma

  =

If there are three washers, the total weight calculates as:

F = 3 * 20 * 9.81

  = 588.6 N

Answer:

ΔL = MmRgt / (2m + M)

Explanation:

The system starts from rest, so the change in angular momentum correlates directly to its final angular momentum.

ΔL = L − L₀

ΔL = Iω − 0

ΔL = ½ MR²ω

To determine the angular velocity ω, begin by drawing a free body diagram for both the pulley and the block.

For the block, two forces act: the weight force mg downward and tension force T upward.

For the pulley, three forces are present: weight force Mg down, a reaction force up, and tension force T downward.

For the sum of forces in the -y direction on the block:

∑F = ma

mg − T = ma

T = mg − ma

For the sum of torques on the pulley:

∑τ = Iα

TR = (½ MR²) (a/R)

T = ½ Ma

Substituting gives:

mg − ma = ½ Ma

2mg − 2ma = Ma

2mg = (2m + M) a

a = 2mg / (2m + M)

The angular acceleration of the pulley is:

αR = 2mg / (2m + M)

α = 2mg / (R (2m + M))

Finally, the angular velocity after time t is:

ω = αt + ω₀

ω = 2mg / (R (2m + M)) t + 0

ω = 2mgt / (R (2m + M))

Substituting into the previous equations gives:

ΔL = ½ MR² × 2mgt / (R (2m + M))

ΔL = MmRgt / (2m + M)

Answer:

Explanation:

The position of the charge q₁ is established at (0,0)

Meanwhile, the charge q₂ is located at (x₁,0)

Thus, the electric potential energy between these two charges is determined by:

Now, the location of charge q₂ shifts from (x₁,0) to (x₂,y₂). The updated electric potential energy between the charges can be represented as:

According to the work-energy theorem, the alteration in potential energy corresponds to the work performed. This is expressed mathematically as:

Consequently, the work done by the electrostatic force on the moving charge is . Therefore, this concludes the solution.

Refer to the diagram below.

Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².

The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s

As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²

Thus, the deceleration magnitude is 82 m/s².