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Crazy boy [7]
3 years ago
14

Which group of components is common to the circulatory systems of most living animals? A. arteries, veins, capillaries B. vessel

s, heart, circulating fluid C. aorta, ventricles, atria D. blood, heart, cavities
Physics
1 answer:
Artyom0805 [142]3 years ago
7 0

The circulatory system, composed of the heart, veins, arteries, capillaries, and blood, is a feature common in most vertebrates (animals with a distinct spine). In these animals, blood never leaves the confines of the heart and blood vessels, making theirs a closed circulatory system. In regard to your question, the best answer is B. Some animals have hearts that don’t have atria, while others don’t even have a distinct heart at all.

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Which one of the following accurately describes the force of gravity?
elena55 [62]
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4 0
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The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
3 years ago
A silver bar 0.125 meter long is subjected to a temperature change from 200 C to 100 C . What will be the length of the bar afte
dimulka [17.4K]
\Delta L= \alpha L_0 (T_f-T_i)

= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)

= -0.00225 m

New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248
So your answer is B.
8 0
3 years ago
Read 2 more answers
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