Answer:
The effective spring constant of the firing mechanism is 1808N/m.
Explanation:
First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:
(This is correct because the horizontal motion has acceleration zero). Then:
Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:
Then, plugging in the given values, we obtain:
Finally, the effective spring constant of the firing mechanism is 1808N/m.
Sí, porque tanto las leyes de Gauss, boyle, etc, relacionan estos comportamientos con el comportamiento de los gases, los cuales son los que ocasionan los diferentes fenómenos naturales y meteorológicos.
Answer:
Explanation:
We can try writing the equation of the horizontal component of the length of the minute hand in terms of distance and the angle, that depends of time in this particular case.
The x-component of the length of the minute hand is:
(1)
- d is the length of the minute hand (d=D/2)
- D is the diameter of the clock
- t is the time (min)
Now, using the angular kinematic equations we can express the angle in term of angular velocity and time. As we know, the minute hand moves with a constant angular velocity, so we can use this equation:
(2)
Also we know, that the minute hand moves 90 degrees or π/2 rad in 15 min, so using the definition of angular velocity, we have:
Now, let's put this value on (2)
Finally the length x(t) of the shadow of the minute hand as a function of time t, will be:
I hope it helps you!
C) fault block, the mountain is forming on a FAULT line.
Yes. B'coz its magnification is always negative.
