Answer:
= 0.517 m
Explanation:
This is a resonance exercise where the ends of the string are fixed, therefore it has a node of them, the fundamental (longest) wavelength created has the form
λ = 2L
wave speed is related to wavelength and frequency
v = λ f
v = 2L f
let's calculate
v = 2 0.50 440
v = 440 m / s
since they indicate that the tension of the string does not change and the linear density of the string is constant, the speed of the wave also remains constant
f =
let's find the length for the new resonance frequency
L = 
let's calculate
L = 
L = 0.5166 m
Answer:
The ratio of moment of inertia of larger sphere to that of smaller sphere = 4
Explanation:
The moment of inertia of solid sphere is given by I = 2/5MR² where M = mass of sphere and R = radius of sphere.
Radius of smaller sphere = D/2
Radius of larger sphere = 2D/2 = D.
Moment of inertia of smaller sphere I₁ = 2/5M × D²/4 = MD²/10
Moment of inertia of larger sphere I₂ = 2/5M × D² = 2MD²/5
The ratio of moment of inertia of larger sphere to that of smaller sphere = I₂/I₁ = 2MD²/5 ÷ MD²/10 = 10 × 2/5 = 4
Answer:
Explanation:
An information contains
25Hz and 75Hz sine wave
Sample frequency is 500Hz
The analogy signal are generally
y(t) = Asin(2πx/λ - wt), w=2πf
y1(t)=Asin(2πx/λ - wt)
y1(t)=Asin(2πx/λ - 2π•25t)
y1(t)=Asin(2πx/λ - 50πt)
Similarly
y2(t)=Asin(2πx/λ - 150πt)
Using Nyquist theorem
Nyquist Theorem states that in order to adequately reproduce a signal it should be periodically sampled at a rate that is 2 times the highest frequency you wish to record.
From sampling
f(nyquist)=f(sample)/2
f(nyquist)=500/2
f(nyquist)=250Hz
From signal
The highest frequency is 150Hz
F(nyquist) = 2×F(highest)
f(nyquist)= 2×150
f(nyquist)= 300Hz
Sample per frequency Ns is given as
Ns=F(sample)/F(highest signal)
Ns=500/150
Ns=3.33sample/period
This is above nyquist rate of 2sample/period
So signal below 300Hz reproduced without aliasing.
The highest resulting frequency is 300Hz
Answer:
B). 3.4 s
Explanation:
As we can see the graph is given between velocity and time
so here we can see that the velocity is changing here with time and initially for some time it moves with constant speed
Then it's speed decreases to next few second and then speed increases to its maximum value
The time after which velocity comes to its maximum value will reach after t = 3 s
so out of the all given options most correct option will be
This is the answer to Question 5
