Answer: no
Explanation: we need a picture
Answer:
a)1815Joules b) 185Joules
Explanation:
Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;
F = ke where;
F is the applied force
k is the elastic constant
e is the extension of the material
From the formula, k = F/e
F1/e1 = F2/e2
If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;
k = 60/0.5
k = 120N/m
a) To get the work done in stretching the spring 5.5m from its position,
Work done by the spring = 1/2ke²
Given k = 120N/m, e = 5.5m
Work done = 1/2×120×5.5²
Work done = 60× 5.5²
Work done = 1815Joules
b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;
Work done = 1/2ke²
Work done =1/2× 120×1.5²
Works done = 60×1.5²
Work done = 135Joules
Answer:
A.−2.1 × 10^10 N
Explanation:
Using the formula;
E = k Q1Q2/d²
Where;
E is the electrical force
k is the constant
Q1, Q2 are the two charges and
d is the distance between the two charges
Therefore;
E = (9 x 10^9) × (0.0042) × (-0.0050) / (0.0030)²
= -2.1 x 10^10 N
Therefore; electrical force acting between the two charges is -2.1 x 10^10 N.
1 atom of Mg(Magnesium) and 2 atoms of Br(Bromine)
