B. so light can shine through it from below.
Answer:
c
Explanation:
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To solve this problem we need to apply the corresponding sound intensity measured from the logarithmic scale. Since in the range of intensities that the human ear can detect without pain there are large differences in the number of figures used on a linear scale, it is usual to use a logarithmic scale. The unit most used in the logarithmic scale is the decibel yes described as

Where,
I = Acoustic intensity in linear scale
= Hearing threshold
The value in decibels is 17dB, then

Using properties of logarithms we have,




Therefore the factor that the intensity of the sound was 
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Answer:
F = 75[J]
Explanation:
We know that work is defined as the product of force by distance.
In this way we have two forces, the weight of the block down, and the force that bring about the block to rise.

where:
W = work = 50 [J]
d = distance = 2 [m]
Fweight = 50 [N]
Fupward [N]
Now replacing:
![50=-(50*2)+(F_{upward}*2)\\50+100=F_{upward}*2\\F_{upward}=150/2\\F_{upward}=75[J]](https://tex.z-dn.net/?f=50%3D-%2850%2A2%29%2B%28F_%7Bupward%7D%2A2%29%5C%5C50%2B100%3DF_%7Bupward%7D%2A2%5C%5CF_%7Bupward%7D%3D150%2F2%5C%5CF_%7Bupward%7D%3D75%5BJ%5D)