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Usimov [2.4K]
3 years ago
8

Noe is studying wave A and wave B. Wave A is rated at 100 dB, and wave B is rated at 90 dB. Which statement can be made about th

e waves
Physics
2 answers:
ololo11 [35]3 years ago
4 0
Wave A is louder than Wave B
IRISSAK [1]3 years ago
4 0

Answer:

Wave a is louder than wave b!!!

Explanation:

Yes

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Sam, whose mass is 72 kg , takes off across level snow on his jet-powered skis. The skis have a thrust of 190 N and a coefficien
Westkost [7]

Answer:

<em>181.11 m</em>

Explanation:

Net Force is the sum of all the resultant force of all the forces acting on a body. The net force experienced by the jet is the difference of the Applied force and frictional force and this is represented in equation 1;

F_{net} = F_{applied} - F_{friction} \\ ............................1

but F_{friction} = μmg .......................................2

Where μ is the coefficient of friction = 0.1

          m is the mass of the body = 72 kg

          g is the acceleration due to gravity 9.81 m/s^{2}

Substituting into equation 2 we have;

F_{friction} = 0.1 x 72 x 9.81

F_{friction} = 70.63 N

Now we substitute our answer in equation 1;

F_{net} = 190 N - 70.63 N \\

F_{net} = 119.37 N

We have to calculate the net acceleration in order to get our velocity .

F_{net} = ma_{net}

a_{net} = \frac{F_{net}}{m}

a_{net}  = \frac{119.37 N}{72 kg }

a_{net}  = 1.66 m/s^{2}

Speed can be express as;

v = u + at ...................3

where u is the initial velocity, which is 0 in this case.

a is the net acceleration  = 1.66 m/s^{2}

t is the time which is 9 s

substituting the values in equation 3 we have

v =  1.66 m/s^{2} x 9 s

v = 14.94 m/s

Calculating for Sam's distance for the first 9 seconds, using the equation of motion we have;

S_{1}  = ut +\frac{1}{2}at^{2}

the jet was initially at rest so initial velocity is 0  and a_{net}  = 1.66 m/s^{2}

S_{1}  = \frac{1}{2} *1.66 m/s^{2} *9^{2}

S_{1} = 67.23 m

Also we have to calculate Sam's distance after nine seconds using equations of motion express below;

v^{2} -u^{2} = 2as

making S the subject formula we have;

S_{2}  =\frac{v^{2}-u^{2}  }{2a} ....................................4

v is the maximum velocity after the fuel finished  and its 14.94 m/s and a is the acceleration along the horizontal plane which put into consideration the coefficient of friction. a = μg = 0.1*9.8 m/s^{2} = 0.98 m/s^{2}

We substitute our values into equation 4 to get our remaining distance;

S_{2}  = \frac{(14.94 m/s)^{2} }{2(0.98 m/s^{2} )}

S_{2} = 113.88 m

Therefore the total distance S = S_{1} + S_{2}

S = 67.23 m + 113.88 m

<em>The total distance covered by Sam is </em>181.11 m

8 0
3 years ago
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