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-Dominant- [34]
3 years ago
6

Homeostasis is the regulation of an organism’s internal environment to maintain conditions suitable for life. Homeostasis requir

es
A) competition between the body’s systems.
B) balance between all of the body’s systems.
C) the body’s systems to act independently of each other.
D) maximum growth of all types of cells and tissues within the body.
Physics
2 answers:
kherson [118]3 years ago
5 0
I think it would be B
serious [3.7K]3 years ago
5 0

Answer:

B) balance between all of the body’s systems.

Explanation:

usatestprep

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A 2.20-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 29.0 N is require
g100num [7]

Answer:

a. 145 N/m b. 1.29 Hz c. 1.62 m/s d.  0 m e. 13.2 m/s² f. ± 0.2 m g. 2.9 J h. 0.54 m/s i. 4.39 m/s²

Explanation:

a. The force constant of the spring

The spring force F = kx and k = F/x where k is the spring constant. F = 29.0 N and x = 0.200 m

k = 29.0 N/0.200 m = 145 N/m

b. The frequency of oscillations, f

f = 1/2π√(k/m)    m = mass = 2.20 kg

f = 1/2π√(145 N/m/2.20 kg) = 1.29 Hz

c. maximum speed of the object

The maximum elastic potential energy of the spring = maximum kinetic  energy of the object

1/2kx² = 1/2mv²

v = (√k/m)x where v is the maximum speed of the object

v = (√145/2.2)0.2 = 1.62 m/s

d Where does the maximum speed occur?

The maximum speed occurs at  0 m

e. The maximum acceleration

a = kx/m = 145 × 0.2/2.2 = 13.2 m/s²

f. The maximum acceleration occurs at x = ± 0.2 m

g. The total energy of the system is the maximum elestic potential energy of the system

E = 1/2kx² = 1/2 × 145 × 0.2² = 2.9 J

h. When x = x₀/3

1/2k(x₀/3)² = 1/2mv²

kx₀²/9 = mv²

v = 1/3(√k/m)x₀ = 1/3(√145/2.2)0.2 = 0.54 m/s

i When x = x₀/3

a = kx₀/3m =  145 × 0.2/(2.2 × 3)= 4.39 m/s²

8 0
3 years ago
In which of the following would the particles move most rapidly? a. ice at -20 °C b. water at 20 °C c. steam at 110 °C d. boilin
Artemon [7]

Answer:

C. steam at 110

Explanation:

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Trick question (really easy) just for fun! If u get this right u get brainliest!
Maurinko [17]

Answer:

piper duh

Explanation:

6 0
2 years ago
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A 0.10 kg mass is oscillating at a small angle from a light string of length 0.10 m.
cupoosta [38]

Answer:

tough

Explanation:

5 0
2 years ago
A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind
Marat540 [252]

Answer:

Part a)

F_v = 4.28 N

Part B)

L = 1.02 m

Part C)

v = 1.25 m/s

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

Tcos\theta = mg

T sin\theta = F_v

\frac{F_v}{mg} = tan\theta

F_v = mg tan\theta

F_v = 1.2\times 9.81 (tan20)

F_v = 4.28 N

Part B)

Here we can use energy theorem to find the distance that it will move

-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2

(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2

(-3.5 + 2.54)L = - 0.98

L = 1.02 m

Part C)

At terminal speed condition we know that

F_v = mg

bv^2 = mg

2.5 v^2 = 3.9

v = 1.25 m/s

7 0
2 years ago
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