Time = 25s
speed = 10m/min
= 10m / 60
= (1/6)m/s
distance = speed × time
= 25 × (1/6)
=4.167m
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:
weaker has the heavier of an object
Spring C stretches 100 cm.
Explanation:
The spring constant is simply the stiffness of the spring. The higher the spring constant the more stiff the spring is.
Spring constant shows the force needed to stretch a spring from it's equilibrium position. If a material requires more force to cause it to stretch, it will have a high spring constant.
According to hooke's law "the force needed to extended an elastic material is directly proportional to its extension"
F = ke
k is the spring constant
e is the extension
We see that the spring that stretches by 100 is the less stiff compared to other springs. It has the smallest spring constant.
Learn more;
Force brainly.com/question/8882476
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