Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.
Explanation:
- Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s
- Let
be the speed of the river's current given as 1.00 m/s.
- Note that this speed is the magnitude of the velocity which is a vector quantity.
- The direction of the swimmer is upstream.
Hence the resultant velocity is given as, 
= S — S 0
= 1.25 — 1
= 0.25 m/s.
Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.
Answer:
Al llegar a su equilibrio térmico ambas barran tendrán una temperatura de 53 grados centígrados.
Explanation:
Dado que una barra de aluminio que está a 78 grados centígrados entra en contacto con una barra de cobre de la misma longitud y área que esta a 28 grados centígrados, y posteriormente se lleva acabo la transferencia de energía entre ambas barras llegando a su equilibrio térmico, para determinar la temperatura a la que ambas barras llegarán se debe realizar el siguiente cálculo:
(78 + 28) / 2 = X
106 / 2 = X
53 = X
Por lo tanto, al llegar a su equilibrio térmico ambas barran tendrán una temperatura de 53 grados centígrados.
(a) 1200 rad/s
The angular acceleration of the rotor is given by:
where we have
is the angular acceleration (negative since the rotor is slowing down)
is the final angular speed
is the initial angular speed
t = 10.0 s is the time interval
Solving for 
, we find the final angular speed after 10.0 s:
(b) 25 s
We can calculate the time needed for the rotor to come to rest, by using again the same formula:
If we re-arrange it for t, we get:
where here we have
is the initial angular speed
is the final angular speed
is the angular acceleration
Solving the equation,
Answer:
If you apply a force to separate 2 opposite poles, the potential energy of the system increases.
Answer:
The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.
Explanation:
Given that,
The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.
The speed of sound in air is 343 m/s.
To find,
The wavelength range for the corresponding frequency.
Solution,
The speed of sound is given by the following relation as :
Wavelength for f = 45 Hz is,
Wavelength for f = 375 Hz is,
So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.
