Answer:
Furthermore, the Pythagorean theorem works when the two added vectors are at right angles to one another - such as for adding a north vector and an east vector.
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
I notice that even though we're working with frames of reference
here, you never said which frame the '5 km/hr' is measured in.
In fact ! You didn't even say which frame the '12 km/hr' of his
bike is measured in.
So there are several different ways this could go. I'll do it the way
I THINK you meant it, but that doesn't guarantee anything.
-- Simon is riding his bike at 12 km/hr relative to the sidewalk,
away from Keesha.
-- He throws a ball at Keesha, at 5 km/hr relative to his own face.
-- Keesha sees the ball approaching her at (12 - 5) = 7 km/hr
relative to the ground and to her.
I’m pretty sure it is an object with a net force of zero. All forces are balanced and EQUAL
Answer:
1. 2.12105 J. The final kinetic energy is. KE f mv2. (875.0 kg)(44.0 m/s)2 2.What is the velocity of the two hockey players after the collision? ... A 10.0-kg test rocket is fired vertically from.
Explanation:
Sana maka tulong