Answer:
a = 3.61[m/s^2]
Explanation:
To find this acceleration we must remember newton's second law which tells us that the total sum of forces is equal to the product of mass by acceleration.
In this case we have:
![F = m*a\\\\m=mass = 3.6[kg]\\F = force = 13[N]\\13 = 3.6*a\\a = 3.61[m/s^2]](https://tex.z-dn.net/?f=F%20%3D%20m%2Aa%5C%5C%5C%5Cm%3Dmass%20%3D%203.6%5Bkg%5D%5C%5CF%20%3D%20force%20%3D%2013%5BN%5D%5C%5C13%20%3D%203.6%2Aa%5C%5Ca%20%3D%203.61%5Bm%2Fs%5E2%5D)
Answer:
The correct answer is B
Explanation:
Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity
Φ
= ∫ E. dA =
/ ε₀
For this case we create a Gaussian surface that is a sphere. We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product
∫ E dA =
/ ε₀
The area of a sphere is
A = 4π r²
E 4π r² =
/ ε₀
E = (1 /4πε₀
) q / r²
Having the solution of the problem let's analyze the points:
A ) r = 3R / 4 = 0.75 R.
In this case there is no charge inside the Gaussian surface therefore the electric field is zero
E = 0
B) r = 5R / 4 = 1.25R
In this case the entire charge is inside the Gaussian surface, the field is
E = (1 /4πε₀
) Q / (1.25R)²
E = (1 /4πε₀
) Q / R2 1 / 1.56²
E₀ = (1 /4π ε₀
) Q / R²
= Eo /1.56
²
= 0.41 Eo
C) r = 2R
All charge inside is inside the Gaussian surface
=(1 /4π ε₀
) Q 1/(2R)²
= (1 /4π ε₀
) q/R² 1/4
= Eo 1/4
= 0.25 Eo
D) False the field changes with distance
The correct answer is B
The acceleration due to gravity on Earth is 9.8 m/s per second.
Answer:
The frequencies are 
Explanation:
From the question we are told that
The length of the ear canal is 
The speed of sound is assumed to be 
Now taking look at a typical ear canal we see that we assume it is a closed pipe
Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

substituting values


Also the the second harmonic for the pipe (ear canal) is mathematically represented as
substituting values
Given that sound would be loudest in the pipe at the frequency, it implies that the child will have an increased audible sensitivity at this frequencies