As part of a carnival game, a 5.00 kg target is freely hanging from a very long and very light wire. Contestants can use one of

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:

The correct option is 80 dB.

Explanation:

The transformation of sound intensity level into sound intensity utilizes the formula

[D] = 10 log (I/I₀)

Where I₀ = 10⁻¹² W/m²

[D] results in 100 dB

100 = 10 log (I/I₀)

Log (I/I₀) converts to 10

(I/I₀) = 10¹⁰

I is determined as I₀ × 10¹⁰ = 10⁻¹² × 10¹⁰ = 10⁻² W/m²

Sound intensity inversely relates to the square of the distance from the source.

I ∝ (1/d²)

I can be expressed as k/d²

When d = 1 m, the intensity is 10⁻² W/m²

Thus, 0.01 = k/1

Providing that k = 0.01 W

For d = 20 m, we can calculate I

I = 0.01/20² = 2.5 × 10⁻⁵ W/m²

With four neighbors mowing their lawns concurrently,

I = 4 × 2.5 × 10⁻⁵ = 10⁻⁴ W/m⁻²

The sound intensity level in decibels is represented as

[D] = 10 log (I/I₀)

[D] = 10 log (10⁻⁴/10⁻¹²)

[D] = 10 log (10⁸)

[D] = 10 × 8 = 80dB

Answer:

31.4 mm²

Explanation:

The ability of a telescope or eye to gather light can be expressed by the formula,

where d signifies the diameter of the pupil.

In bright daylight, the usual size of the pupil is 3 mm.

Conversely, in darkness, the diameter typically enlarges to 7 mm.

This indicates an increase in light-gathering capacity.

Thus, the amount of light the eye can capture is 31.4 mm².

Answer:

Speeds of 1.83 m/s and 6.83 m/s

Explanation:

Based on the law of conservation of momentum,

where m represents mass, is the initial speed before impact, and are the velocities of the impacted object after the collision and of the originally stationary object after the impact.

Thus,

After the collision, the kinetic energy doubles, therefore:

Substituting the initial velocity of 5 m/s provides the equation needed to proceed.

We know that leads to

Using the quadratic formula leads us to solve for the speeds after the explosion, specifically where a=2, b=-10, and c=-25.

By substituting the values, the solution yields results for the speeds of the blocks, which are ultimately 1.83 m/s and 6.83 m/s.