Answer:

Explanation:
Given data
Force F=2 N
Length L=17 cm = 0.17 m
Spring Constant k=42 N/m
To find
Relaxed length of the spring
Solution
From Hooke's Law we know that

Answer:
(a) 0.063 m/s
(b) 1.01 m/s
Explanation:
rate of volume flow, V = 4 x 10^-6 m^3/s
(a) radius, r = 4.5 x 10^-3 m
Let the speed of blood is v.
So, V = A x v
where A be the area of crossection of artery
4 x 10^-6 = 3.14 x 4.5 x 10^-3 x 4.5 x 10^-3 x v
v = 0.063 m/s
Thus, the speed of flow of blood is 0.063 m/s .
(b) Now r' = r / 4 = 4.5 /4 x 10^-3 m = 1.125 x 10^-3 m
Let the speed is v'.
So, V = A' x v'
4 x 10^-6 = 3.14 x 1.125 x 10^-3 x 1.125 x 10^-3 x v'
v' = 1.01 m/s
Thus, the speed of flow of blood is 1.01 m/s .
Answer:
D
Explanation:
The power equation is P= V^2/R
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Answer:
1.9 m.
Explanation:
Three complete waves in the length of 5.7 m
The distance traveled by one complete wave is called wavelength.
Thus, the distance traveled by one wave = 5.7 / 3 = 1.9 m
Thus, the wavelength is 1.9 m.
Complete question:
Consider the hypothetical reaction 4A + 2B → C + 3D
Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?
Answer:
the final concentration of A is 0.992 M.
Explanation:
Given;
time of reaction, t = 4.0 s
rate of change of the concentration of B = -0.0760 M/s
initial concentration of A = 1.600 M
⇒Determine the rate of change of the concentration of A.
From the given reaction: 4A + 2B → C + 3D
2 moles of B ---------------> 4 moles of A
-0.0760 M/s of B -----------> x

⇒Determine the change in concentration of A after 4s;
ΔA = -0.152 M/s x 4s
ΔA = -0.608 M
⇒ Determine the final concentration of A after 4s
A = A₀ + ΔA
A = 1.6 M + (-0.608 M)
A = 1.6 M - 0.608 M
A = 0.992 M
Therefore, the final concentration of A is 0.992 M.