Which planets have rings
2 answers:
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The weight of the meterstick is:
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
For counting x you use simple equation for the distance covered by the object when it moves with constant velocity:
that gives you 20m after 1st second, 40 m after 2nd second, 60 m after 3rd second and so on.
For counting y you have to use the equation for the distanced covered by the object moving with constantly accelerating velocity (symbols refering to vertical movement):
that gives you 5m after 1st second, 20m afters 2nd second, 45m after 3rd second and so on.
Add minus signs before y positions to receive graph presenting the movement of the ball.
So the points are: P1=[20,-5], P2=[40,-20], P3=[60,-45] and so on... Pn=[x,y].
that gives you 20m after 1st second, 40 m after 2nd second, 60 m after 3rd second and so on.
For counting y you have to use the equation for the distanced covered by the object moving with constantly accelerating velocity (symbols refering to vertical movement):
that gives you 5m after 1st second, 20m afters 2nd second, 45m after 3rd second and so on.
Add minus signs before y positions to receive graph presenting the movement of the ball.
So the points are: P1=[20,-5], P2=[40,-20], P3=[60,-45] and so on... Pn=[x,y].
To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.
The altitude is,
And the velocity can be written as,
From the properties of standard atmosphere at altitude z = 20km temperature is
Velocity of sound at this altitude is
Then the Mach number
So front stagnation temperature
Therefore the temperature at its front stagnation point is 689.87K