Aplicación de las palancas en la Fisioterapia.
1 answer:
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The available options are: (found the complete text on internet)
A- at a distance less than r
B- at a distance equal to r
C- at a distance greater than r
Solution:
The correct answer is C) at a distance greater than r.
In fact, the gravitational attraction between the satellite and the Earth provides the centripetal force that keeps the satellite in circular orbit, so we can write
where the term on the left is the gravitational force, while the term on the right is the centripetal force, and where
G is the gravitational constant
M is the Earth mass
m is the satellite mass
r is the distance of the satellite from the Earth's center
v is the satellite speed
Re-arranging the equation, we get
and we see from this formula that, if the second satellite has a speed less than the speed v of the first satellite, it means that the denominator of the fraction is smaller, and so r is larger for the second satellite.
A- at a distance less than r
B- at a distance equal to r
C- at a distance greater than r
Solution:
The correct answer is C) at a distance greater than r.
In fact, the gravitational attraction between the satellite and the Earth provides the centripetal force that keeps the satellite in circular orbit, so we can write
where the term on the left is the gravitational force, while the term on the right is the centripetal force, and where
G is the gravitational constant
M is the Earth mass
m is the satellite mass
r is the distance of the satellite from the Earth's center
v is the satellite speed
Re-arranging the equation, we get
and we see from this formula that, if the second satellite has a speed less than the speed v of the first satellite, it means that the denominator of the fraction is smaller, and so r is larger for the second satellite.
Answer:
Second drop: 1.04 m
First drop: 1.66 m
Explanation:
Assuming the droplets are not affected by aerodynamic drag.
They are in free fall, affected only by gravity.
I set a frame of reference with the origin at the nozzle and the positive X axis pointing down.
We can use the equation for position under constant acceleration.
X(t) = x0 + v0 * t + 1/2 * a *t^2
x0 = 0
a = 9.81 m/s^2
v0 = 0
Then:
X(t) = 4.9 * t^2
The drop will hit the floor when X(t) = 1.9
1.9 = 4.9 * t^2
t^2 = 1.9 / 4.9
That is the moment when the 4th drop begins falling.
Assuming they fall at constant interval,
Δt = 0.62 / 3 = 0.2 s (approximately)
The second drop will be at:
X2(0.62) = 4.9 * (0.62 - 1*0.2)^2 = 0.86 m
And the third at:
X3(0.62) = 4.9 * (0.62 - 2*0.2)^2 = 0.24 m
The positions are:
1.9 - 0.86 = 1.04 m
1.9 - 0.24 = 1.66 m
above the floor