All categories

3 years ago

What life process requires specialized cells in multicellular organisms

1 answer:

4 0

The most probable answer for this question would be that almost every life process requires specialized cells in multicellular organisms. To simply put it, cells of multicellular organisms are specialized in a way that they are all grouped into their respective tissues and these tissues are all grouped into their respective organs and these organs are all grouped together into their respective systems and these systems make up the multicellular organisms. These systems have their own functions in maintaining and sustaining the life that the organisms has. The organs have their own functions as well, thus specialized cells are mostly needed in respiration, digestion, circulation, movement, excretion, reproduction, immunity, coordination, and synthesis.

You might be interested in

At takeoff, an aircraft travels at 62 m/s, so that the air speed relative to the bottom of the wing is 62 m/s. Given the sea lev

olganol [36]

Answer:

the aircraft must travel at a speed of 73.4 m/s in order to create the ideal lift.

Explanation:

We will use Bernoulli's theorem in order to determine the pressure lift:

ΔP = 1/2 (ρ)(v₂² - v₁²)

the generated pressure lift is ΔP = 1000 N/m²

Therefore,

1000 = 1/2(ρ)(v₂² - v₁²)

v₂² - v₁² = 2000 / ρ

v₂² = (2000 N/m² / 1.29 kg/m³) + (62 m/s)²

v₂ = √[ (2000 N/m² / 1.29 kg/m³) + (62 m/s)² ]

v₂ = 73.4 m/s

Therefore, the aircraft must travel at a speed of 73.4 m/s in order to create the ideal lift.

5 0

3 years ago

particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine

AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is $1062.7\frac{m}{s^{2}}$

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

$F=ma$ (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

$F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}$

with q1 and q2 the charge of the particles, r the distance between them and k the constant $k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}$ . So:

$F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}$

$F=62.7 N$

Using that value on (1) and solving for a

$a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}$

4 0

3 years ago

The potential energy of a 1.7 x 10-3 kg particle is described by U (x )space equals space minus (17 space J )space cos open squa

hichkok12 [17]

Answer:

Explanation:

Given a particle of mass

M = 1.7 × 10^-3 kg

Given a potential as a function of x

U(x) = -17 J Cos[x/0.35 m]

U(x) = -17 Cos(x/0.35)

Angular frequency at x = 0

Let find the force at x = 0

F = dU/dx

F = -17 × -Sin(x/0.35) / 0.35

F = 48.57 Sin(x/0.35)

At x = 0

Sin(0) =0

Then,

F = 0 N

So, from hooke's law

F = -kx

Then,

0 = -kx

This shows that k = 0

Then, angular frequency can be calculated using

ω = √(k/m)

So, since k = 0 at x = 0

Then,

ω = √0/m

ω = √0

ω = 0 rad/s

So, the angular frequency is 0 rad/s

4 0

3 years ago

Do all metals stick to magnets? Give an example. I NEED HELP FASTT

ludmilkaskok [199]

Answer:

Explanation:

Not all metals stick to magnets. Like aluminum. if you were to stick a magnet on to an aluminum it would fall off.

6 0

2 years ago

2- A student ran 135 meters in 15 seconds. What was the student's velocity?

RoseWind [281]

Answer:

9 Brainly hahaha ............huh

4 0

2 years ago