Answer:
The minimum frequency is 702.22 Hz
Explanation:
The two speakers are adjusted as attached in the figure. From the given data we know that
=3m
=4m
By Pythagoras theorem
Now
The intensity at O when both speakers are on is given by
Here
- I is the intensity at O when both speakers are on which is given as 6
- I1 is the intensity of one speaker on which is 6
- δ is the Path difference which is given as
- λ is wavelength which is given as
Here
v is the speed of sound which is 320 m/s.
f is the frequency of the sound which is to be calculated.
where k=0,1,2
for minimum frequency 
, k=1
So the minimum frequency is 702.22 Hz
Answer:
C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂
Explanation:
You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.
Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.
Stone 1
y₁ = v₀₁ t + ½ g t²
y₁ = 0 + ½ g t²
Rock2
It comes out a little later, let's say a second later, we can use the same stopwatch
t ’= (t-t₀)
y₂ = v₀₂ t ’+ ½ g t’²
y₂ = 0 + ½ g (t-t₀)²
y₂ = + ½ g (t-t₀)²
Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to
S = y₁ -y₂
S = ½ g t²– ½ g (t-t₀)²
S = ½ g [t² - (t²- 2 t to + to²)]
S = ½ g (2 t t₀ - t₀²)
S = ½ g t₀ (2 t -t₀)
This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.
For t <to. The rock y has not left and the distance increases
For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time
passes
Now we can analyze the different statements
A) false. The difference in height increases over time
B) False S increases
C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂
Hi there!
We can use the work-energy theorem to solve.
Recall that:
The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:
Now, we can define work:
Now, plug in the values:
Solve for theta:
Answer:
Tension in the string is equal to 58.33 N ( this will be the strength of the string )
Explanation:
We have given mass m = 1.7 kg
radius of the circle r = 0.48 m
Kinetic energy is given 14 J
Kinetic energy is equal to 
So 
v = 4.05 m/sec
Centripetal force is equal to
So tension in the string will be equal to 58.33 N ( this will be the strength of the string )
