When a 5 kg rock is dropped from a height of 6 m on Planet X, it loses 24 J of GPE. What is the acceleration due to gravity on P
1 answer:
Answer:
g = 1.25m/s²
Explanation:
Given the following data;
Mass = 5kg
Height = 6m
Gravitational potential energy = 24J
To find the acceleration due to gravity;
Potential energy can be defined as an energy possessed by an object or body due to its position.
Mathematically, potential energy is given by the formula;
Where,
P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
GPE = mgh
Substituting into the equation, we have;
24 = 5*6*g
24 = 30g
g = 30/24
g = 1.25m/s²
Therefore, the acceleration due to gravity on Planet X is 1.25m/s².
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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.
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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
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m is the satellite mass
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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.