Consider a coin which is tossed straight up into the air. After it is released it moves upward, reaches its highest point and fa
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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:
a= - 6.667 m/s²
Explanation:
Given that
The initial speed of the box ,u= 20 m/s
The final speed of the box ,v= 0 m/s
The distance cover by box ,s= 30 m
Lets take the acceleration of the box = a
We know that
v²= u ² + 2 a s
Now by putting the values in the above equation we get
0²=20² + 2 a x 30
a= - 6.667 m/s²
Negative sign indicates that velocity and acceleration are in opposite direction.
Therefore the acceleration of the box will be - 6.667 m/s² .
Answer:
The value is
Explanation:
From the question we are told that
The magnitude of the horizontal force is
The mass of the crate is
The acceleration of the crate is
Generally the net force acting on the crate is mathematically represented as
Here is force of kinetic friction (in N) acting on the crate
So
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