Answer:
Explanation:
Given data:
Weight W = 16 lb
length l = 6/12 = 0.5 ft
hence, spring constant k = W/l = 16/0.5 = 32 lb/ft
The equation of motion of spring is
the auxiliary equation can be written as
The discriminate of equation is
To get the value of the damping constant,
Answer:
Explanation:
Relative to ground level it has
PE = mgh = 8(10)(5) = 400 J
Answer:
Object D
Explanation:
Use Newton's Second Law to determine the acceleration that each object has.
The force applied in both cases is 50 N, but the mass for object C and object D is different.
Let's start with object C first:
- F = ma
- 50 N = 10 kg · a
- 50 = 10a
- 5 = a
The acceleration object C undergoes is 5 m/s².
Now let's calculate object D next:
- F = ma
- 50 N = 2 kg * a
- 50 = 2a
- 25 = a
The acceleration object D undergoes is 25 m/s².
Object D has greater acceleration because it has a smaller mass. The object with a smaller mass will accelerate more in order to satisfy Newton's 2nd Law.
Answer:
Ф = 239.73 rad
Explanation:
α = 12 + 15×t
W = ∫α×dt
= ∫(12 + 5×t)×dt
= 12×t + 2.5×t^2
then:
Ф = ∫W×dt
= ∫(12×t + 2.5×t^2)dt
= 6×t^2 + 5/6×t^3
therefore the angle at t = 4.88s is:
Ф = 6×(4.88)^2 + 5/6×(4.88)^3
= 239.73 rad
Its A: the use of hydropower often changes the natural flow of water through an ecosystem
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