Answer:The place to go for the answer to such an easy question is the SI Brochure, the document which defines the SI and all its units.
Answer: ME= E total - E thermal
Answer:
The intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².
Explanation:
Given;
intensity of the sound level, dB = 60 dB
The intensity of the sound in W/m² is calculated as;
![dB = 10 Log[\frac{I}{I_o} ]\\\\](https://tex.z-dn.net/?f=dB%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C)
where;
I₀ is threshold of hearing = 1 x 10⁻¹² W/m²
I is intensity of the sound in W/m²
Substitute the given values and for I;
![dB = 10 Log[\frac{I}{I_o} ]\\\\60 = 10 Log[\frac{I}{I_o} ]\\\\6 = Log[\frac{I}{I_o} ]\\\\10^6 = \frac{I}{I_o} \\\\I = 10^6 \ \times \ I_o\\\\I = 10^6 \ \times \ 1^{-12} \ W/m^2 \\\\I = 1\ \times \ 10^{-6} \ W/m^2](https://tex.z-dn.net/?f=dB%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C60%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C6%20%3D%20%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C10%5E6%20%3D%20%5Cfrac%7BI%7D%7BI_o%7D%20%5C%5C%5C%5CI%20%3D%2010%5E6%20%5C%20%5Ctimes%20%5C%20I_o%5C%5C%5C%5CI%20%3D%2010%5E6%20%5C%20%5Ctimes%20%5C%201%5E%7B-12%7D%20%5C%20W%2Fm%5E2%20%5C%5C%5C%5CI%20%3D%201%5C%20%5Ctimes%20%5C%2010%5E%7B-6%7D%20%5C%20W%2Fm%5E2)
Therefore, the intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².
Acceleration = (change in speed) / (time for the change) = 9/3 = <em>3 m/s²</em> .
His mass makes no difference.
Answer:
Explanation:
Given that,
At one instant,
Center of mass is at 2m
Xcm = 2m
And velocity =5•i m/s
One of the particle is at the origin
M1=? X1 =0
The other has a mass M2=0.1kg
And it is at rest at position X2= 8m
a. Center of mass is given as
Xcm = (M1•X1 + M2•X2) / (M1+M2)
2 = (M1×0 + 0.1×8) /(M1 + 0.1)
2 = (0+ 0.8) /(M1 + 0.1)
Cross multiply
2(M1+0.1) = 0.8
2M1 + 0.2 =0.8
2M1 = 0.8-0.2
2M1 = 0.6
M1 = 0.6/2
M1 = 0.3kg
b. Total momentum, this is an inelastic collision and it momentum after collision is given as
P= (M1+M2)V
P = (0.3+0.1)×5•i
P = 0.4 × 5•i
P = 2 •i kgm/s
c. Velocity of particle at origin
Using conversation of momentum
Momentum before collision is equal to momentum after collision
P(before) = M1 • V1 + M2 • V2
We are told that M2 is initially at rest, then, V2=0
So, P(before) = 0.3V1
We already got P(after) = 2 •i kgm/s in part b of the question
Then,
P(before) = P(after)
0.3V1 = 2 •i
V1 = 2/0.3 •i
V1 = 6 ⅔ •i m/s
V1 = 6.667 •i m/s
