a surface recieving sound is moved from it original position to a position three times farther away from the source of the sound
1 answer:
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Answer:
(a) Force must be grater than 283.87 N
(B) Force will be equal to 193.945 N
Explanation:
We have given mass of the crate m = 49.6 kg
Acceleration due to gravity
Coefficient of static friction
Coefficient of kinetic friction
(a) Static friction force is given by
So to just start the crate moving we have to apply more force than 283.87 N
(B) This force will be equal to kinetic friction force
We know that kinetic friction force is given by
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.