The spring has been stretched 0.701 m
Explanation:
The elastic potential energy of a spring is the potential energy stored in the spring due to its compression/stretching. It is calculated as
where
k is the spring constant
x is the elongation of the spring with respect to its equilibrium position
For the spring in this problem, we have:
E = 84.08 J (potential energy)
k = 342.25 N/m (spring constant)
Therefore, its elongation is:
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That depends on how much of it you have, but it's a hugely enormous gigantic number in even the smallest sample.
You can see from the formula that there are 12 atoms in every molecule of the stuff !
Answer:
Looks like you have:
a = -.324 cos 2.5 t
In this case ω^2 A = .324
ω = 2.5
f = ω / (2 * pi) = 2.5 / 6.28 = .40 / sec
Answer:
Explanation:
We shall represent all the displacement in terms of vector . We shall consider east as i , and north as j . south of west direction will be represented by - i - j .
displacement of 1.9 km due east
D₁ = 1.9 i
vector representing south of west = - i - j
unit vector = - i - j / √ 2
7.2 km south of west = 7.2 ( - i - j ) / √ 2
D₂ = - 5.09 ( i + j )
Total displacement
= D₁ + D₂
= 1.9 i - 5.09 ( I + J )
D = - 3.19 i - 5.09 j .
magnitude of D = √ ( 3.19² + 5.09² )
= 6 km .
Direction of D
Tanθ = 5.09 / 3.19 = 1.59
θ = 58°
So direction will be 58° south of west .
To reach the starting point , he shall have to go in opposite direction .
So he shall have to go in the direction of north of east at angle 58° by a displacement of 6 km .
