Two charges separated by one meter exert a 9 N force on each other. If the charges are pushed to a 3 meter separation, the force
1 answer:
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the force that the planet exerts on the moon is equal to the force that the moon exerts on the planet
Explanation:
In this problem we are analzying the gravitational force acting between a planet and its moon.
The magnitude of the gravitational attraction between two objects is given by
where :
is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between them
In this problem, we are considering a planet and its moon. According to Newton's third law of motion,
"When an object A exerts a force (action force) on an object B, then object B exerts an equal and opposite force (reaction force) on object A"
If we apply this law to this situation, this means that the force that the planet exerts on the moon is equal to the force that the moon exerts on the planet.
Learn more about gravitational force:
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There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill:
- radius of the hill:
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car
(downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, 
, so we can write:
(1)
By rearranging the equation and substituting the numbers, we find N:
(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
from which we find
- speed of the car at the top of the hill:
- radius of the hill:
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car
By rearranging the equation and substituting the numbers, we find N:
(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
from which we find
Complete Question
The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .
What eyepiece focal length will give the microscope an overall angular magnification of 300?
Answer:
The eyepiece focal length is
Explanation:
From the question we are told that
The focal length is
This negative sign shows the the microscope is diverging light
The angular magnification is
The distance between the objective and the eyepieces lenses is
Generally the magnification is mathematically represented as
Where is the eyepiece focal length of the microscope
Now making the subject of the formula
substituting values