All categories

3 years ago

Which of the falling animals filter feeds?

Physics

2 answers:

Gelneren [198K]3 years ago

7 0

The answer is that all marine animals feed by seperating food particles from water, known as filter feeding.

Sveta_85 [38]3 years ago

3 0

The most common examples of filter-feeding are clams and sponges.

Filter feeding is a process in whch animals, normally animals that can't move fast enough to catch pray and haven't evolved to do so, separate useful particles from the water in order to feed.

Hope it helped,

BioTeacher101

You might be interested in

A permanent magnet is pushed into a wire, left there for a while, and then pulled out. During which time does a current run thou

lakkis [162]

Answer:

C. while the magnet is moving

Explanation:

Electromagnetic induction implies the production of electric current by mere movement of a magnet with respect to a coil or wire.

In the given question, current would be induced in the wire only when the magnet moves. That is either when the magnet is pushed into a wire, or when pulled out. But no current would flow through the wire when the magnet is left there for a while.

The current is induced because of the motion involved. Thus, the appropriate option is C.

8 0

3 years ago

Based on its type of chemical bond, which of the following has the highest boiling point?

Sever21 [200]

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

Potassium chloride

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

Potassium Chloride is an ionic compound and has ionic bond which is stronger than covalent bond in ethyl alcohol,water ,ammonia, and thus has the highest boiling point .
ionic bond is a type of bond that results from the transfer of electrons between metallic atoms and non-metallic atoms.

5 0

2 years ago

A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

I am Lyosha [343]

Answer:

Rate of change of magnetic field is $3.466\times 10^3T/sec$

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop $r=\frac{0.13}{2}=0.065m$

Length of the circular loop $L=2\pi r=2\times 3.14\times 0.065=0.4082m$

Wire is made up of diameter of 2.6 mm

So radius $r=\frac{2.6}{2}=1.3mm=0.0013m$

Cross sectional area of wire $A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2$

Resistivity of wire $\rho =2.18\times 10^{-8}m$

Resistance of wire $R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm$

Current is given i = 11 A

So emf $e=11\times 1.67\times 10^{-3}=0.0183volt$

Emf induced in the coil is $e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}$

$0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}$

$\frac{dB}{dt}=3.466\times 10^3=T/sec$

8 0

3 years ago

A heavy anvil is suspended by a 0.75 m long steel wire that has a mass of 12 g. When the wire is plucked, it hums at its fundame

Dima020 [189]

Explanation:

It is given that,

length of steel wire, l = 0.75 m

Mass of the wire, m = 12 g = 0.012 kg

Fundamental frequency, f = 120 Hz

We need to find the mass of the anvil (m'). The fundamental frequency is given by :

$f=\dfrac{v}{2l}$

v is the speed of the mass

Speed is given by :

$v=\sqrt{\dfrac{T}{\mu}}$

$\mu$ is the mass per unit length, $\mu=\dfrac{m}{l}$

$f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}$

T is the tension in the wire,

$f=\dfrac{1}{2l}\sqrt{\dfrac{Tl}{m}}$

$T=4f^2lm$

$T=4(120)^2\times 0.75\times 0.012$

T = 518.4 N

Tension in the wire, T = m' g

$m'=\dfrac{T}{g}$

$m'=\dfrac{518.4}{9.8}$

m' = 52.89 kg

So, the mass of the anvil is 52.89 kg. Hence, this is the required solution.

6 0

3 years ago

Two identical twins hold on to a rope, one at each end, on a smooth, frictionless ice surface. They skate in a circle about the

777dan777 [17]

Answer:

Part a)

$L = 2683.2 kg m^2/s$

Part b)

$v' = 8.60 m/s$

Part c)

$W = 4326.7 J$

Explanation:

Part a)

As we know that there is no external torque on the system of two twins

so here we will use

$L = mv r + mvr$

$L = 2(78 \times 4.30 \times 4)$

$L = 2683.2 kg m^2/s$

Part b)

Since angular momentum is conserved here as there is no external torque

so we will have

$2(m v r) = 2( m v' \frac{r}{2})$

$v' = 2v$

$v' = 8.60 m/s$

Part c)

Work done by both of them = change in kinetic energy

so we have

$W = 2(\frac{1}{2}mv'^2 - \frac{1}{2}mv^2)$

$W = m(v'^2 - v^2)$

$W = 78(8.60^2 - 4.3^2)$

$W = 4326.7 J$

5 0

2 years ago