Answer:
x = 6.94 m
Explanation:
For this exercise we can find the speed at the bottom of the ramp using energy conservation
Starting point. Higher
Em₀ = K + U = ½ m v₀² + m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
Let's calculate
v = √(1.23² + 2 9.8 1.69)
v = 5.89 m / s
In the horizontal part we can use the relationship between work and the variation of kinetic energy
W = ΔK
-fr x = 0- ½ m v²
Newton's second law
N- W = 0
The equation for the friction is
fr = μ N
fr = μ m g
We replace
μ m g x = ½ m v²
x = v² / 2μ g
Let's calculate
x = 5.89² / (2 0.255 9.8)
x = 6.94 m
2.89watts.
<h3>What is meant by sound intensity?</h3>
- The average rate at which sound energy moves across a unit area normal to a given direction is used to determine a sound's intensity. This rate is generally stated in ergs per second per square centimeter.
- Decibels are the units used to measure sound intensity, often known as sound power or sound pressure. The decibel (dB) unit is named after Alexander Graham Bell, who also created the audiometer and the telephone. An audiometer is a tool to gauge a person's hearing capacity for various noises.
- Our ability to measure the flow of sound energy as a time-averaged vector quantity makes sound intensity measuring an effective method. We can identify sound sources and tell direct sound from reverberant sound in a room using the characteristics of sound intensity.
How much power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m:
Formula: 
I=1.6x10-3 w/m2
r=12m




To learn more about sound intensity, refer to:
brainly.com/question/17062836
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Answer:
If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface.
Explanation:
Option A is incorrect because, given this case, it is easier to calculate the field.
Option B is incorrect because, in a situation where the surface is placed inside a uniform field, option B is violated
Option C is also incorrect because it is possible to be a field from outside charges, but there will be an absence of net flux through the surface from these.
Hence, option D is the correct answer. "If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface."