Question 7 (2 points)
1 answer:
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Refer to the diagram shown below.
The given data is
mass, kg Coordinates. m
------------- -----------------
2 (0, 0)
2 (2, 0)
4 (2, 1)
Total mass, M = 2+2+4 = 8kg
Let (x,y) be the coordinates of M.
Then, taking moments about the origin, we obtain
8x = 2*0 + 2*2 + 4*2 = 12
x = 1.5 m
8y = 2*0 + 2*0 + 4*1 = 4
y = 0.5 m
Answer: (1.5, 0.5) m
The given data is
mass, kg Coordinates. m
------------- -----------------
2 (0, 0)
2 (2, 0)
4 (2, 1)
Total mass, M = 2+2+4 = 8kg
Let (x,y) be the coordinates of M.
Then, taking moments about the origin, we obtain
8x = 2*0 + 2*2 + 4*2 = 12
x = 1.5 m
8y = 2*0 + 2*0 + 4*1 = 4
y = 0.5 m
Answer: (1.5, 0.5) m
Answer:
Explanation:
Given data
Electric potential at point a is Ua=5.4×10⁻⁸J
q₂ moves to point b where a negative work done on it
Required
Electric potential energy Ub
Solution
When a particle moves from a point where the potential is Ua to a point where it is Ub the change in potential energy is equal to work done where the force exerted on the charge is conservative and work done is given by:
Now substitute the given values
So