3.278*10^6 I think. Sorry if it’s wrong.
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:
proof in explanation
Explanation:
First, we will calculate the number of half-lives:

where,
n = no. of half-lives = ?
t = total time passed = 2100 million years
= half-life = 700 million years
Therefore,

Now, we will calculate the number of uranium nuclei left (
):

and the rest of the uranium nuclei will become thorium nuclei (
)

dividing both:

<u>Hence, it is proven that after 2100 million years there are seven times more thorium nuclei than uranium nuclei in the rock.</u>
Time required : 3 s
<h3>Further explanation
</h3>
Power is the work done/second.

To do 33 J of work with 11 W of power
P = 11 W
W = 33 J

Answer: 996m/s
Explanation:
Formula for calculating velocity of wave in a stretched string is
V = √T/M where;
V is the velocity of wave
T is tension
M is the mass per unit length of the wire(m/L)
Since the second wire is twice as far apart as the first, it will be L2 = 2L1
Let V1 and V2 be the speed of the shorter and longer wire respectively
V1 = √T/M1... 1
V2 = √T/M2... 2
Since V1 = 249m/s, M1 = m/L1 M2 = m/L2 = m/2L1
The equations will now become
249 = √T/(m/L1) ... 3
V2 = √T/(m/2L1)... 4
From 3,
249² = TL1/m...5
From 4,
V2²= 2TL1/m... 6
Dividing equation 5 by 6 we have;
249²/V2² = TL1/m×m/2TL1
{249/V2}² = 1/2
249/V2 = (1/2)²
249/V2 = 1/4
V2 = 249×4
V2 = 996m/s
Therefore the speed of the wave on the longer wire is 996m/s