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.

You might be interested in

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x''

Lady_Fox [76]

Answer:

We know that the acceleration of the particle is defined as

$a(t)=\frac{dv}{dt}$

Since it is given that

$v(t)=\frac{5}{\sqrt{t}}\\\\\therefore a(t)=\frac{d(\frac{5}{\sqrt{t}})}{dt}\\\\a(t)=5\times \frac{dt^{-\frac{1}{2}}}{dt}\\\\=\frac{-5}{2}t^{\frac{-1}{2}-1}\\\\\therefore a(t)=x''(t)=\frac{-2.5}{t^{\frac{3}{2}}}$

Now by definition of velocity we have

$v(t)=\frac{dx(t)}{dt}\\\\\Rightarrow dx(t)=v(t)dt$

Integrating on both sides we get

$v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int v(t)dt$

Applying values we get

$v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int \frac{5}{t^{\frac{1}{2}}}dt\\\\\therefore x(t)=\frac{5}{0.5}\int t^{-0.5}dt\\\\x(t)=10\sqrt{t}+c$

To find the constant we note that at t=1 the particle is at x=12 Thus applying values in the above equation we get

$c=12-10\\\therefore c=2\\\\x(t)=10\sqrt{t}+2$

6 0

3 years ago

A railroad cart with a mass of m1 = 11.6 t is at rest at the top of an h = 10.9 m high hump yard hill.

leonid [27]

The final common speed of the two carts will be 69.3 m/sec.The momentum conservation principle is applied.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

Unit conversion:

1 metric tons = 1000 kg

Given data;

m₁= 11.6 metric ton =11600 kg

m₂ = 23.2 metric ton = 23200 kg

Let v represent the combined velocity of the two carts once they are connected, and let u represent the starting velocity of cart 1 when it reaches the bottom.

Considering energy conservation;

$\rm m_1 g h = \frac{1}{2} m_1 \times u^2 \\\\ u^2 = 2gh\\\\ u^2 = 2 \times 9.8 \times 10.6 \\\\ u = 207.972 \ m/s$