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3 years ago

Which of the following is an echinoderm that moves by waving its arms?

Physics

2 answers:

charle [14.2K]3 years ago

6 0

The answer to this question is b

sukhopar [10]3 years ago

4 0

B. Brittle Star is an echinoderm that moves by waving its arms.

Brittle star is a spiny, hard-skinned, long-armed animal, related to sea star, that lives on rocky sea floor, from shallow areas to the deepest areas. It uses its flexible arms for locomotion.

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The valence electrons of metals are weakly attracted to the parent nuclei, so the electrons break free and float. The moving ele

siniylev [52]

The valence electrons of metals are weakly attracted to the parent nuclei, so the electrons break free and float. The moving electrons form a electron <u>negative</u> blanket that binds the atomic <u>positive</u> nuclei together, forming a metallic bond.

So the answers are <u>{ Negative }</u> and <u>{ Positive }.</u>

Please vote Brainliest (:

5 0

3 years ago

Read 2 more answers

2. A can filled with sand has a mass of 0.65kg is swung overhead in a horizontal circle of radius 0.70m at a constant rate of 2.

Aliun [14]

<h3><u>Answer</u>;</h3>

≈ 5 Kgm²/sec

<h3><u>Explanation</u>;</h3>

Angular momentum is given by the formula

L = Iω, where I is the moment of inertia and ω is the angular speed.

I = mr², where m is the mass and r is the radius

= 0.65 × 0.7²

= 0.3185

Angular speed, ω = v/r

= (2 × 3.142 × r × 2.5) r

= 15.71

Therefore;

Angular momentum = Iω

= 0.3185 × 15.71

= 5.003635

<u>≈ 5 Kgm²/sec</u>

6 0

2 years ago

Two students are on a balcony 19.1 m above the street. One student throws a ball, b1, vertically downward at 13.9 m/s. At the sa

tester [92]

Answer:

Part a)

$t = 2.83 s$

Part b)

Ball thrown downwards = $v_f = 23.8 m/s$

Ball thrown upwards = $v_f = 23.8 m/s$

Part c)

$d = 22.24 m$

Explanation:

Part a)

Since both the balls are projected with same speed in opposite directions

So here the time difference is the time for which the ball projected upward will move up and come back at the same point of projection

Afterwards the motion will be same as the first ball which is projected downwards

so here the time difference is given as

$\Delta y = 0 = v_y t + \frac{1}{2}at^2$

$0 = 13.9 t - \frac{1}{2}(9.81) t^2$

$t = 2.83 s$

Part b)

Since the displacement in y direction for two balls is same as well as the the initial speed is also same so final speed is also same for both the balls

so it is given as

$v_f^2 - v_i^2 = 2 a \Delta y$

$v_f^2 - (13.9)^2 = (2)(-9.81)(-19.1)$

$v_f^2 = 567.9$

$v_f = 23.8 m/s$

Part c)

Relative speed of two balls is given as

$v_{12} = v_1 - v_2$

$v_{12} = (13.9) - (-13.9) = 27.8 m/s$

now the distance between two balls in 0.8 s is given as

$d = v_{12} t$

$d = 27.8 \times 0.8$

$d = 22.24 m$

7 0

3 years ago

How can you tell that this paragraph uses a quotation?<br> Check all that apply?

Debora [2.8K]

Answer- There are two reasons that we know quotations have been used first is the use of of name of the person who quoted it and secondly the quotation is written inside the quotation marks.

Explanation- Quotation is nothing but using a line that has been already quoted by someone somewhere. Such sentences are normally written inside quotation marks. In the above given paragraph the name of the person who quotes the sentence is also given hence we know that our quotation has been used.

8 0

3 years ago

The mass of jupiter is 300 times the mass of the earth. Jupiter orbits the sun with Tjupiter = 11.9 yr in an orbit with Rjupiter

mihalych1998 [28]

Answer:

c) 11.9 yr

Explanation:

The orbital period is proportional to r^(3/2) and does not depend on the satellite's mass. Any object at Jupiter position will have the same orbital period regardless of mass.

By keppler's law we know that

T^2= r^3

T= orbital time period

r= mean distance of the planet from the Sun.

clearly, The orbital period does not depend on the satellite's mass

there, the correct answer will be c= 11.9 yr.

5 0

3 years ago