Answer:
281.25 J
Explanation:
We are told that the two objects with masses m and 3m.
Also that energy stored in the spring is 375 joules.
Now, initially the centre of mass of the system took place at rest, it means v1 = v and v2 = v/3
Thus, from principle of conservation of energy, we have;
½mv² + ½(3m)(v/3)² = 375J
(m + 3m/9)½v² = 375
(4/3)m × ½v² = 375
Multiply both sides by ¾ to get;
½mv² = 375 × ¾
½mv² = 281.25 J
Therefore, energy of lighter body is 281.25 J
Better weight distribution and more stability
Answer:
<h2>
d₂ = 3d</h2><h2>
The diameter of the second wire is 3 times that of the initial wire.</h2>
Explanation:
Using the formula for calculating the resistivity of an object to find the diameter.
Resistivity P = RA/L
R is the resistance of the material
A is the cross sectional area
L is the length of the material
Since A = πd²/4
P = R( πd²/4)/L
P = Rπd²/4L ... 1
If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;
P₂ = (R/9)A₂/L₂
P₂ = (R/9)(πd₂²/4)/L₂
P₂ = (Rπd₂²/36)/L₂
P₂ = (Rπd₂²)/36L₂
Since the length and resistivity are the same;
P = P₂ and L =L₂
Equating 1 and 2;
Rπd²/4L = (Rπd₂²)/36L₂
Rπd²/4L = (Rπd₂²)/36L
d² = d₂²/9
d₂² = 9d²
Taking the square root of both sides;
√d₂² = √9d²
d₂ = 3d
Therefore the diameter of the second wire is 3 times that of the initial wire
