Answer:
Vibrations of the eardrum are in turn translated into oscillations of tiny bones (ossicles) found within the middle ear. The Hammer, Anvil, and Stirrup magnify the oscillations and relay this information to the chamber of the inner ear.
Answer:
So the sound intensity level they would experience without the earplugs is 110.32dB.
Explanation:
Given data
Sound intensity by factor =215
Sound intensity level =87 dB
To find
Sound intensity level they would experience without the earplugs
Solution
First we need to find the new sound intensity level
So
The dB can be calculated as:
Substitute the given values
So the sound intensity level they would experience without the earplugs is 110.32dB.
Are you asking about independent and dependent variables?
<h2>
Answer:</h2>
D. (1m, 0.5m)
<h2>
Explanation:</h2>
The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;
x = (m₁x₁ + m₂x₂ + m₃x₃) / M ----------------(i)
y = (m₁y₁ + m₂y₂ + m₃y₃) / M ----------------(ii)
Where;
M = sum of the masses
m₁ and x₁ = mass and position of first mass in the x direction.
m₂ and x₂ = mass and position of second mass in the x direction.
m₃ and x₃ = mass and position of third mass in the x direction.
y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.
From the question;
m₁ = 6kg
m₂ = 4kg
m₃ = 2kg
x₁ = 0m
x₂ = 3m
x₃ = 0m
y₁ = 0m
y₂ = 0m
y₃ = 3m
M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg
Substitute these values into equations (i) and (ii) as follows;
x = ((6x0) + (4x3) + (2x0)) / 12
x = 12 / 12
x = 1 m
y = (6x0) + (4x0) + (2x3)) / 12
y = 6 / 12
y = 0.5m
Therefore, the center of mass of the system is at (1m, 0.5m)
1.5 m/s is the velocity.
9.3 m is the length of aisle, over which Distance will be covered.
Time is demanded in which the child will move the cart over the aisle with 1.5 m/s.
v=S/t
and,
t=S/v
Put values,
t=9.3/1.5=6.2 s
