Answer: First, we determine the circumference of the Mars by the equation below.
C = 2πr
Substituting the known values,
C = 2(π)(3,397 km) = 6794π km
To determine the tangential speed, we divide the circumference calculated above by the time it takes for Mars to complete one rotation and that is,
tangential speed = 6794π km / 24.6 hours = 867.64 km/h
First we write the corresponding kinematics equations:
a = -g
v = -g * t + vo
y = -g * ((t ^ 2) / 2) + vo * t + yo
Substituting the values:
y = - (9.81) * (((0.50) ^ 2) / 2) + (19) * (0.50) + (0) = 8.27m
answer:
the displacement at the time of 0.50s is 8.27m
Answer:
F = W + ma a> 0
Explanation:
For this exercise let's use Newton's second law
we assume the upward direction as positive
F - W = m a
F = W + ma
F = m (g + a)
In this case they indicate that the speed is less and less as it goes down, therefore the acceleration must be opposite to the speed, that is, the acceleration is upwards, consequently it is positive
We can see that since a> 0 the force F must have greater than the weight of the elevator
<h2>
Answer:</h2>
In circuits, the average power is defined as the average of the instantaneous power over one period. The instantaneous power can be found as:
So the average power is:
But:
So:
![P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}(\frac{1+cos(2\omega t)}{2} )dt \\\\P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}[\frac{1}{2}+\frac{cos(2\omega t)}{2}]dt \\\\P=\frac{v_{m}i_{m}}{T}[\frac{1}{2}(t)\right|_0^T +\frac{sin(2\omega t)}{4\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2T}[(t)\right|_0^T +\frac{sin(2\omega t)}{2\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7BT%7D%5Cintop_%7B0%7D%5E%7BT%7D%28%5Cfrac%7B1%2Bcos%282%5Comega%20t%29%7D%7B2%7D%20%29dt%20%5C%5C%5C%5CP%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7BT%7D%5Cintop_%7B0%7D%5E%7BT%7D%5B%5Cfrac%7B1%7D%7B2%7D%2B%5Cfrac%7Bcos%282%5Comega%20t%29%7D%7B2%7D%5Ddt%20%5C%5C%5C%5CP%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7BT%7D%5B%5Cfrac%7B1%7D%7B2%7D%28t%29%5Cright%7C_0%5ET%20%2B%5Cfrac%7Bsin%282%5Comega%20t%29%7D%7B4%5Comega%7D%20%5Cright%7C_0%5ET%5D%20%5C%5C%20%5C%5C%20P%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7B2T%7D%5B%28t%29%5Cright%7C_0%5ET%20%2B%5Cfrac%7Bsin%282%5Comega%20t%29%7D%7B2%5Comega%7D%20%5Cright%7C_0%5ET%5D%20%5C%5C%20%5C%5C%20P%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7B2%7D)
In terms of RMS values:
