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2 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.

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A 70.0 kg man jumping from a window lands in an elevated fire rescue net 11.0 m below the window. He momentarily stops when he h

levacccp [35]

Answer:

70.15 Joule

Explanation:

mass of man, m = 70 kg

intial length, l = 11 m

extension, Δl = 1.5 m

Let K is the spring constant.

In the equilibrium position

mg = K l

70 x 9.8 = K x 11

K = 62.36 N/m

Potential energy stored, U = 0.5 x K x Δl²

U = 0.5 x 62.36 x 1.5 x 1.5

U = 70.15 Joule

6 0

2 years ago

Web Spiders and Oscillations All spiders have special organs that make them exquisitely sensitive to vibrations. Web spiders det

chubhunter [2.5K]

Complete Question

The complete quetion is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

The mass of the fly is $m_f = 11 mg = 11*10^{-3} g = 1.1*10^{-5} \ kg$

The extension of the web is $e= 4.00 \ mm = 0.004 \ m$

The spring constant is mathematically evaluated as

$k = \frac{mg}{e}$

substituting values

$k = \frac{1.1 *10^{-5} *9.8}{0.004}$

$k = 0.027 \ N/m$

The frequency of vibration is

$f = \frac{1}{2 \pi} \sqrt{\frac{k}{m} }$

substituting values

$f = \frac{1}{2 * 3.142 } \sqrt{\frac{0.027}{1.1*10^{-5}} }$

$f = 7.9 Hz$

8 0

2 years ago

Help with 1 2 and 3 please

geniusboy [140]

1. Amperes, is the SI unit (also a fundamental unit) responsible for current.

2. $I = \frac{q}{t}$ Δq over Δt technically

Rearrange for Δq

I x Δt = Δq

1.5mA x 5 = Δq

Δq = 0.0075

Divide this by the fundamental charge "e"

Electrons: 0.0075 / 1.60 x 10^-19

Electrons: 4.6875 x 10^16 or 4.7 x 10^16

3. So we know that the end resistances will be equal so:

ρ = RA/L

ρL = RA

ρL/A = R

Now we can set up two equations one for the resistance of the aluminum bar and one for the copper: Where 1 represents aluminum and 2 represents copper

$\frac{p1L1}{A1} = \frac{p2L2}{A2}\\$