Answer:
Option B
Explanation:
The phase difference is found by subtracting the 2.3m for the receiver from the other speaker which is 2.9m hence
Phase difference= 2.9-2.3= 0.6
Answer:
<em>Entropy Change = 0.559 Times</em>
Explanation:
Entropy change is determined by the change in the micro-states of a system. As we know that the micro-states are the same as measure of disorderness between initial and final states, that's the the amount of change in micro-states determine how much of entropy has changed in the system.
Answer:
Explanation:
i = Imax sin2πft
given i = 180 , Imax = 200 , f = 50 , t = ?
Put the give values in the equation above
180 = 200 sin 2πft
sin 2πft = .9
sin2π x 50t = .9
sin 360 x 50 t = sin ( 360n + 64 )
360 x 50 t = 360n + 64
360 x 50 t = 64 , ( putting n = 0 for least value of t )
18000 t = 64
t = 3.55 ms .
Answer:

Explanation:
During the exchange of applied force, thermal energy is generated by the friction that exists between the ground and the tire.
Said force according to the statement is the reaction of half the force on the rear tire. In this way the normal force acted is,

The work done is given by the friction force and the distance traveled,

Where ![\mu_k [/ tex] is the coefficient of kinetic frictionN is the normal force previously found d is the distance traveled,Replacing,[tex]W_f = (0.80)(441)(0.42)](https://tex.z-dn.net/?f=%20%5Cmu_k%20%5B%2F%20tex%5D%20is%20the%20coefficient%20of%20kinetic%20friction%3C%2Fp%3E%3Cp%3EN%20is%20the%20normal%20force%20previously%20found%20d%20is%20the%20distance%20traveled%2C%3C%2Fp%3E%3Cp%3EReplacing%2C%3C%2Fp%3E%3Cp%3E%5Btex%5DW_f%20%3D%20%280.80%29%28441%29%280.42%29)
The thermal energy released through the work done is,

Answer: a) 456.66 s ; b) 564.3 m
Explanation: The time spend to cover any distance a constant velocity is given by:
v= distance/time so t=distance/v
The slower student time is: t=780m/0.9 m/s= 866.66 s
For the faster students t=780 m/1,9 m/s= 410.52 s
Therefore the time difference is 866.66-410.52= 456.14 s
In order to calculate the distance that faster student should walk
to arrive 5,5 m before that slower student, we consider the follow expressions:
distance =vslower*time1
distance= vfaster*time 2
The time difference is 5.5 m that is equal to 330 s
replacing in the above expression we have
time 1= 627 s
time2 = 297 s
The distance traveled is 564,3 m