Answer:
3.16 × 
W/
Explanation:
β(dB)=10 × 
=
W/
β=55 dB
Therefore plugging into the equation the values,
55=10 ![log_{10}(\frac{I}{ [tex]10^{-12}](https://tex.z-dn.net/?f=log_%7B10%7D%28%5Cfrac%7BI%7D%7B%20%5Btex%5D10%5E%7B-12%7D)
})[/tex]
5.5= })[/tex]
= 
316227.76×= I
I= 3.16 × W/
The distance is 30 km and the displacement is 22.4 km North East
Answer:
268N
Explanation:
The upward force acting on the block are the reaction and the hooked table..
The total normal force acting = normal reaction + 24N
Note that the normal reaction is always equal the weight of the table
Hence the normal force acting in the block is 244.0+24 = 268.0N
Answer:
B) 100 J
Explanation:
Assuming the distance given is measured along the incline, the vertical change in height is (5 m)(sin 30°) = 2.5 m. Then the change in potential energy is ...
∆PE = mg(∆h) = (4 kg)(10 m/s^2)(2.5 m) = 100 J
