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2 years ago

During development, the cells in these organisms become specialized.

Physics

1 answer:

Y_Kistochka [10]2 years ago

7 0

Answer:

A- multicellular

Explanation:

fertilized egg I think that is the answer

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Please help on this one

Yuki888 [10]

The answer is c because I had the same thing and it was right

5 0

3 years ago

A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s

Leona [35]

Answer:

0.6

Explanation:

The volume of a sphere = $\frac{4}{3} \pi (\frac{D}{2})^3$

Therefore $\pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3$

r of the disc = $1.15(\frac{ D}{2} )$

Using conservation of angular momentum;

The $M_i$ of the sphere = $\frac{2}{5} m \frac{D}{2}^2$

$M_i$ of the disc = $m*\frac{ \frac{1.15*D}{2}^2 }{2}$

$\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{ m * \frac{(\frac{`.`5*D}{2})^2 }{2} }$

= 0.6

5 0

2 years ago

A 0.30-kg object is traveling to the right (in the positive direction) with a speed of 3.0 m/s. After a 0.20 s collision, the ob

andre [41]

Answer:

$I = -2.1 kg.m/s$

Explanation:

Given,

mass of the object,m = 0.30 Kg

initial speed, v_i = 3 m/s

time of collision = 0.20 s

final speed, v_f = -4 m/s

Impulse = change in momentum

$I = m (v_f-v_i)$

$I = 0.30\times (-4-3)$

$I = -2.1 kg.m/s$

Hence, impulse of the object is equal to $I = -2.1 kg.m/s$

5 0

2 years ago

Which Richter magnitude range can be recorded by instruments but isn't felt? A. less than 2.9 B. 3.0 – 4.9 C. 5.0 – 5.9 D. 6.0 a

algol [13]

A. anything less than 3.0 magnitude on a richters scale usually can't be felt by humans but instruments can pick it up.

3 0

3 years ago

Read 2 more answers

A bug on the surface of a pond is observed to move up and down a total vertical distance of 6.5 cm , from the lowest to the high

m_a_m_a [10]

Answer:

factor that bug maximum KE change is 0.52284

Explanation:

given data

vertical distance = 6.5 cm

ripples decrease to = 4.7 cm

solution

We apply here formula for the KE of particle that executes the simple harmonic motion that is express as

KE = (0.5) × m × A² × ω² .................1

and kinetic energy is directly proportional to square of the amplitude.

$\frac{KE2}{KE1} = \frac{A2^2}{A1^2}$ .............2

$\frac{KE2}{KE1} = \frac{4.7^2}{6.5^2}$

$\frac{KE2}{KE1}$ = 0.52284

so factor that bug maximum KE change is 0.52284

5 0

3 years ago