An object is realsed From rest . how far does it fall during the second second of fall?
1 answer:
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The distance of the sound from the tuba is 4.82 m.
<h3>Area of the tube</h3>
The area of the tuba is calculated as follows;
I = P/A
where;
- I is intensity of sound
- P is power
- A is area
A = P/I
A = 0.35 / (1.2 x 10⁻³)
A = 291.67 m²
<h3>Distance of the sound</h3>Area = 4πr²
Thus, the distance of the sound from the tuba is 4.82 m.
Learn more about intensity of sound here: brainly.com/question/4431819
Answer:
a) 20.29N
b) 19.43N
c) 15N
Explanation:
To find the magnitude of the resultant vectors you first calculate the components of the vector for the angle in between them, next, you sum the x and y component, and finally, you calculate the magnitude.
In all these calculations you can asume that one of the vectors coincides with the x-axis.
a)
b)
c)
Answer:
222.2N/m
Explanation:
Given parameters:
Elastic potential energy = 4000J
Extension = 6m
Unknown:
Spring coefficient = ?
Solution:
The elastic potential energy is the energy stored within a string.
It is expressed as;
EPE = k e²
k is the spring constant
e is the extension
4000 = x k x 6²
8000 = 36k
k = 222.2N/m