Answer:
x₁ = 0.1878 m
Explanation:
For this exercise we will use conservation of energy
starting point. Highest point
Em₀ = U = m g h
final point. Lowest point with fully compressed spring
Em_f = K_e + U
Em_f = ½ K x² + m g x
energy is conserved
Em₀ = Em_f
m g h = ½ K x² + m g x
½ K x² + mg (x- h) = 0
let's substitute
½ 7.3 x² + 0.030 9.8 (x- 0.25) = 0
3.65 x² + 0.294 (x- 0.25) = 0
x² + 0.080548 (x- 0.25) = 0
x² - 0.020137 + 0.080548 x = 0
x² + 0.080548 x - 0.020137 = 0
let's solve the quadratic equation
x = [0.080548 ±√ (0.080548² + 4 0.020137)] / 2
x = [0.080548 ± 0.29502] / 2
x₁ = 0.1878 m
x₂ = -0.1072 m
These are the compression and extension displacement of the spring
The answer is B. Force=mass X accleration
Answer:liquid water
Explanation:it turns into liquid water when it rains
Answer:
patient receiving drug 25 MCG/minute
Explanation:
given data
infusing = 15 ml/hr
drug = 50 mg
D5W = 500 ml
to find out
How many MCG/minute
solution
we know infusing rate is 15 ml/hr = 0.25 ml/min
so 0.25 ml drug content = 50 /500 × 0.25
0.25 ml drug content = 0.025 mg
so here
rate of drug will be 0.025 mg
rate of drug = 0.025 mg = 25 × gm/min
rate of drug = 25 MCG/minute
so patient receiving drug 25 MCG/minute
Answer:
I think you ask for another way of writing the joule (J). It can be expressed as the product between a force applied to a body and the deployment of the body (not necessarily caused by that force applied), so N*m (newton * meter).