A disk is spinning about its center with a constant angular speed at first. Let the turntable spin faster and faster, with const
ant angular acceleration ?. Which sketch qualitatively shows the direction of the acceleration vector a of the bug?
1 answer:
Answer:
4 (please see the attached file)
Explanation:
While the angular speed (counterclockwise) remained constant, the angular acceleration was just zero.
So, the only force acting on the bug (parallel to the surface) was the centripetal force, producing a centripetal acceleration directed towards the center of the disk.
When the turntable started to spin faster and faster, this caused a change in the angular speed, represented by the appearance of an angular acceleration α.
This acceleration is related with the tangential acceleration, by this expression:
at = α*r
This acceleration, tangent to the disk (aiming in the same direction of the movement, which is counterclockwise, as showed in the pictures) adds vectorially with the centripetal force, giving a resultant like the one showed in the sketch Nº 4.
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Answer:
.
Explanation:
This problem can be solved using Newton's law of universal gravitation: ,
where F is the gravitational force between two masses and
,
is the distance between the masses (their center of mass), and
is the gravitational constant.
We know the weight of the astronout on the surface, with this we can find his mass. Letting be the weight on the surface:
,
,
,
since we now that we get that the mass is
.
Now we can use Newton's law of universal gravitation
,
where is the mass of the astronaut and
is the mass of the earth. From Newton's second law we know that
,
in this case the acceleration is the gravity so
, (<u>becarefull, gravity at this point is no longer</u>
<u>because we are not in the surface anymore</u>)
and this get us to
, where
is his new weight.
We need to remember that the mass of the earth is and its radius is
.
The total distance between the astronaut and the earth is
meters.
Now we can compute his weigh:
,
,
.
We know that acceleration is change in velocity by time taken for that change.
In this case velocity change is 3.7 m/s
Time taken for this change = 60 ms =
So acceleration of frog =
= 61.66 m/
So acceleration of frog is 61.66 m/
o it is evident that frog is capable of remarkable accelerations.