Answer:
Option as B is correct At the same speed as before
Explanation:
As we know the relation between speed of the wave and tension in string
The speed of wave in stretched string
ν = 
speed of wave is the directly proportional to the square root of tension as mentioned in question tension of string is unaffected when in linear mass density is constant, so we can say that the speed of wave will be the same
Option as B is correct At the same speed as before
Answer:
limitations of the measuring instrument.
Explanation:
Hope this helps
Answer:
<h2><em>
6000 counts per second</em></h2>
Explanation:
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;
2000 counts per second = 1 meter ... 1
In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;
x count per second = 3 meter ... 2
Solving the two expressions simultaneously for x we will have;
2000 counts per second = 1 meter
x counts per second = 3 meter
Cross multiply to get x
2000 * 3 = 1* x
6000 = x
<em></em>
<em>This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample</em>
<h2>
Answer:</h2>
1000th multiple of the standard reference level for intensities.
<h2>
Explanation:</h2>
The sound intensity level (β), measured in decibels, of a sound with an intensity of I is defined as follows;
β = 10 log (I / I₀) --------------------(i)
Where;
I₀ = reference intensity
Given from the question;
β = sound level = 30dB
Substitute this value into equation (i) as follows;
30 = 10 log (I / I₀)
Divide both sides by 3;
3 = log (I / I₀)
Take antilog of both sides;
10^(3) = (I / I₀)
1000 = I / I₀
Solve for I;
I = 1000I₀
Therefore the intensity of the sound is 1000 times the standard reference level for intensities (I₀)
