a mass spring system makes 50 complete osillations in 10 seconds. what is the period and frequency of the oscillations
1 answer:
You might be interested in
Complete Question
The compete question is shown on the first uploaded question
Answer:
The speed is
Explanation:
From the question we are told that
The distance of separation is d = 4.00 m
The distance of the listener to the center between the speakers is I = 5.00 m
The change in the distance of the speaker is by
The frequency of both speakers is
Generally the distance of the listener to the first speaker is mathematically represented as
Generally the distance of the listener to second speaker at its new position is
Generally the path difference between the speakers is mathematically represented as
Here is the wavelength which is mathematically represented as
=>
=>
=>
Here n is the order of the maxima with value of n = 1 this because we are considering two adjacent waves
=>
=>
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
Answer:
The energy that the truck lose to air resistance per hour is 87.47MJ
Explanation:
To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by
Our values are:
Replacing,
We need calculate now the energy lost through a time T, then,
But we know that d is equal to
Where
v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,
(per hour)
Therefore the energy that the truck lose to air resistance per hour is 87.47MJ