An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 43.8 N,
1 answer:
You might be interested in
1)
The force of gravitation between the two objects is given by:
where
is the gravitational constant
m1 = 20,000 kg is the mass of the first object
m2 = 12,500 kg is the mass of the second object
r = 5 m is the distance between the two objects
Substituting the numbers inside the equation, we find
2)
From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:
this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor
So, the gravitational force increases by a factor 4. Therefore, the new force will be
3) 12.5 Nm
The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:
Where, in this case:
F = 25 N is the perpendicular force
d = 0.5 m is the distance between the force and the center
By using the equation, we find
4) 0.049 kg m^2/s
The relationship between angular momentum (L), moment of inertia (I) and angular velocity () is:
In this problem, we have
So, the angular momentum is