slader A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of

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

The height measures 69.68 m

Explanation:

Given data

Before striking the ground =  46 % of the total distance

To establish

the height

Solution

We know here acceleration and displacement, which is

d = (0.5)gt²..............1

Here d is the distance, g is the acceleration, and t is time

So, when an object falls it will be

h = 4.9 t²....................2

For the first part of the inquiry

The falling objects account for

54 % of the total distance

0.54 h = 4.9 (t-1)²...................3

Thus,

Now we possess two equations with unknown variables

We can equate both equations

The first equation already solves for h

Substituting h in the second equation allows us to find t

0.54 × 4.9 t² = 4.9 (t-1)²  

t = 0.576 s and  3.771 s

We choose here 3.771 s since 0.576 s is negligible; the distance covered in the last second before it impacts the ground is 46 % of the entire fall.

Thus, selecting t = 3.771 s

Then h from equation 2

h = 4.9 t²

h = 4.9 (3.771)²

h =  69.68 m

Thus, the height is 69.68 m

Stopping distance.

Flow rate calculations yield 220 cans, each with a volume of 0.355 l, leading to 78.1 l/min or 1.3 l/s or 0.0013 m³/s.

At Point 2:
A2 = 8 cm² = 0.0008 m²
V2 = Flow rate/A2 = 0.0013/0.0008 = 1.625 m/s
P1 = 152 kPa = 152000 Pa

At Point 1:
A1 = 2 cm² = 0.0002 m²
V1 = Flow rate/A1 = 0.0013/0.0002 = 6.5 m/s
P1 =?
Height = 1.35 m

Using Bernoulli’s principle;
P2 + 1/2 * V2² / density = P1 + 1/2 * V1² / density + density * gravitational acceleration * height
=> 152000 + 0.5 * (1.625)² * 1000 = P1 + 0.5 * (6.5)² * 1000 + (1000 * 9.81 * 1.35)
=> 153320.31 = P1 + 34368.5
=> P1 = 1533210.31 - 34368.5 = 118951.81 Pa = 118.95 kPa

O momento total, p = 21.24 kg-m/s