Answer:If an object's speed changes, or if it changes the direction it's moving in,
then there must be forces acting on it. There is no other way for any of
these things to happen.
Once in a while, there may be a group of forces (two or more) acting on
an object, and the group of forces may turn out to be "balanced". When
that happens, the object's speed will remain constant, and ... if the speed
is not zero ... it will continue moving in a straight line. In that case, it's not
possible to tell by looking at it whether there are any forces acting on it
We are given with the expression d = ut + 0.5 at^2 and is asked to express the equation in terms of a. First, we transpose ut to the left side, then we multiply to the equation and divide lastly the resulting equation by t^2. The final expression becomes a = 2(d-ut)/t^2.
A) 3 x 10 ^ 8
b) 3 x 10 ^ 5
c) 3.2 x 10 ^ 7
d) 9.6 x 10 ^ 15 m
e) 9.6 x 10 ^ 17 cm
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
