Answer:
formula used K=F/∆l
∆l is the elongation of the spring
- F=10N
- ∆l=20mm===> 0.02m
- K=10N divided 0.02m= 500N/m
Answer: part a: 19.62m
part b: 19.62 m/s
part a: 2.83 secs
Explanation:If the air resistance is ignored then the swimmer experience free fall under gravity hence
u=0
a=9.81 m/s2
t=2 secs
s=h
Part b
Part c
now we have h=2*19.62=39.24
<span>A+B-C
</span><span>A = 6x - 2y
B = -4x - 8y
C = -3x + 9y
(</span>6x - 2y) + (-4x - 8y) - (-3x + 9y)
(6x - 2y) + (-4x - 8y) + (3x - 9y)
2x -10y + (3x - 9y)
5x - 19y
Answer:
10.6cm
Explanation:
We are given 5.3cm below the starting point (spring extension).
Therefore, to find static vertical equilibrium, we use the equation:
kx = mg
Where:
k = spring constant =
=mg/5.3 kg/s²
We are told the object was dropped from rest.
Therefore:
loss in potential energy = gain in spring p.e
Let's use the expression:
mgx = ½kx²
We are asked to find the stretch at maximum elongation x.
To find x, we make x subject of the formula.
Therefore, we have:
x = 2mg/k (after rearranging the equation above)
x = (2mg) / (mg/5.3)
x = 10.6cm
The difference between an sa and sb galaxy is Sa has a larger nucleus and more tightly wound arms than an Sb.
<h3>What is sa Galaxy?</h3>
Sa galaxies are normal spirals that have narrow, tightly wound arms, which usually are visible because of the presence of interstellar dust and, in many cases, bright stars.
<h3>What is sb Galaxy?</h3>
The arms of sb galaxy are more widely spread than those of the Sa variety and appear less smooth.
Thus, the difference between an sa and sb galaxy is Sa has a larger nucleus and more tightly wound arms than an Sb.
Learn more about sa galaxy here: brainly.com/question/9813915
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