A meter stick is pivoted at the 0.50-m line. A 3.0-kg object is hung from the 0.15-m line. Where should a 5.0-kg object be hung
to achieve equilibrium (the meter stick oriented horizontal and motionless)?
1 answer:
Answer:
0.71 m
Explanation:
Let the 5 kg is hang at a distance d from the midpoint.
use the concept of moments.
take moments about the mid point.
Anticlockwise moments = clockwise moments
3 x (0.5 - 0.15) = 5 x d
1.05 = 5 d
d = 0.21 m
So, the distance of 5 kg from 0 mark on the stick is 0.5 + 0.21 = 0.71 m
You might be interested in
Answer:
The centripetal force acting on the skater is <u>48.32 N.</u>
Explanation:
Given:
Radius of circular track is,
Tangential speed of the skater is,
Mass of the skater is,
We are asked to find the centripetal force acting on the skater.
We know that, when an object is under circular motion, the force acting on the object is directly proportional to the mass and square of tangential speed and inversely proportional to the radius of the circular path. This force is called centripetal force.
Centripetal force acting on the skater is given as:
Now, plug in the given values of the known quantities and solve for centripetal force, . This gives,
Therefore, the centripetal force acting on the skater is 48.32 N.
Answer:
The value is
Explanation:
From the question we are told that
The Coulomb constant is
The charge on the electron/proton is
The mass of proton
The mass of electron is
Generally for the electron to be held up by the force gravity
Then
Electric force on the electron = The gravitational Force
i.e