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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

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An archer pulls her bowstring back 0.376 m by exerting a force that increases uniformly from zero to 251 N. (a) What is the equi

Amanda [17]

Answer:

(A) 667.5 N/m

(B)

Explanation:

(A) Let the spring constant be k.

Using the formula F = kx

k = 251 / 0.376

K = 667.5 N/m

(B)

Work done

W = 0.5 × kx^2

W = 0.5 × 667.5 × 0.376 × 0.376

W = 47.2 J

8 0

3 years ago

The length of a rectangle is four times its width. The perimeter of the rectangle is at most 130 cm. Which inequality models the

leva [86]

The answer would be the one where the left side is less than or equal to 130 because 130 has to be the greatest value and the value of the w's cannot exceed 130, but they can equal it

4 0

3 years ago

Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the sp

kotegsom [21]

Answer:

m2 = 83.3 g

Explanation:

by conservation of momentum principle we have

$m_1v_{i1} + m_2v_{i2} = m_2v_{f2}$

as both sphere has same speed so $v_{i2} = v_{i1}$

$m_2 = \frac{m_1}[\frac{v_f2}{v_{f1}}+1}$

from conservation of kinetic energy principle we have

$\frac{1}{2}m_1v^{2}_{i1} + \frac{1}{2}m_2v^{2}_{i2} = \frac{1}{2}m_2v^{2}_f2$

$v_{f1} = \sqrt {\frac{(m_1+m_2) v^2_i1}{m_2}$

$v_{f1} = v_{i2}\sqrt {\frac{(m_1+m_2)}{m_2}$

$\frac{v_{f1}}{v_{i2}} =\sqrt {\frac{(m_1+m_2)}{m_2}$

substituting this value in above equation to get m2 value

$m_2 = \frac{m_1}{\sqrt {\frac{(m_1+m_2)}{m_2}+1}}$

solving for m2 we get

$m2 = \frac{m_1}{3}$

m_1 = 250 g

$=\frac{250}{3}$

m2 = 83.3 g

7 0

3 years ago

Can someone please help me with these physics problems? I just don’t even know where to start.

KIM [24]

for the block of mass 5 kg normal force is given as

$F_n = mg$

$F_n = 5*9.8 = 49 N$

friction force is given as

$F_f = \mu F_n$

$F_f = 0.1*49 = 4.9 N$

Net force is given as

$F_{net} = ma$

$F_{net} = 5*2 = 10 N$

now we know that

$F_{net} = F_{app} - F_f$

$10 = F_{app} - 4.9$

$F_{app} = 14.9 N$

Normal force is given as

$F_n = mg$

$F_n = 6*9.8$

$F_n = 58.8 N$

now we know that

$F_{net} = F_{app} - F_f$

$F_{net} = 0$

as object moves with constant velocity

$F_{app} = F_f = 15 N$

now for coefficient of friction we can use

$F_f = \mu F_n$

$15 = \mu * 58.8$

$\mu = 0.255$

net force upwards is given as

$F = 1.2 * 10^{-4} N$

mass is given as

$m = 7 * 10^{-5} kg$

now as per newton's law we can say

$F = ma$

$1.2 * 10^{-4} = 7 * 10^{-5} * a$

$a = 1.71 m/s^2$

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

$F_{net} = F_{app} - F_f$

$ma = F_{app} - F_f$

$5*6 = 40 - F_{f}$

$F_f = 40 - 30 = 10 N$

part b)

for coefficient of friction we can use

$F_f = \mu F_n$

$10 = \mu * F_n$

here normal force is given as

$F_n = mg = 5*9.8 = 49 N$

now we have

$\mu = \frac{10}{49} = 0.204$

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$20 = 0 + \frac{1}{2}a(5^2)$

$a = 1.6 m/s^2$

now net force is given as

$F_{net} = ma$

$F_{net} = 10*1.6 = 16 N$

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

$a = \frac{v_f - v_i}{t}$

$a = \frac{0 - 25}{1.5} = -16.67 m/s^2$

now the force is gievn as

$F = ma$

$F = 5*16.67 = 83.3 N$

3 0

4 years ago

What is the reactive force when a fish swims through water?

natima [27]

The answer is D

This is an example of Newton's third law of motion.

The active force is the fish against the water, so the reactive force would be the reverse, the equal force of the water on the fish.

7 0

3 years ago