Ask question

Ask question

All categories

notsponge [240]

3 years ago

5

What patterns did you notice in the organism's embryonic development?

1 answer:

kap26 [50]3 years ago

6 0

Germinal stage. 1.1 Fertilization. 1.2 Cleavage. 1.3 Blastulation. 1.4 Implantation. 1.5 Embryonic disc.
Gastrulation.
Neurulation.
Development of organs and organ systems.

You might be interested in

A squirrel on a limb near the top of a tree loses its grip on a nut, so that the nut slips away horizontally at a speed of 1.5m/

tatiyna

Answer:

The height of the squirrel $h =113 \ m$

Explanation:

From the question we are told that

The horizontally speed is $v = 1.5 m/s$

The horizontal distance $d = 7.2 \ m$

Generally the the time taken is mathematically represented

$t = \frac{d}{v}$

=> $t = \frac{7.2}{1.5}$

=> $t = 4.8 \ s$

Generally from kinematic equation we have that

$h = ut + \frac{1}{2} gt^2$

Here u is the initial velocity of nut in the vertical and the value is 0 m/s

$h = 0 * 4.8 + \frac{1}{2} * 9.8 * (4.8)^2$

=> $h =113 \ m$

6 0

3 years ago

The time required from the passing of one crest to the next is called the wave's

Strike441 [17]

That's the wave's ' period '.
It's the reciprocal of the wave's frequency.

8 0

3 years ago

Which of these binary systems is most likely to contain a black hole?A.An x-ray binary containing an O star and another object o

STALIN [3.7K]

Brain signals are converted

5 0

3 years ago

Gravity is the strongest non-contact force on Earth.<br><br> a. true <br> b. false

Semmy [17]

Answer:

$a. trueeeeeeeeee........$

6 0

3 years ago

Read 2 more answers

A neutron star is an extremely dense, rapidly spinning object that results from the collapse of a massive star at the end ofits

Westkost [7]

Answer:

(a). The rotational inertia is $5.72\times10^{39}\ kg m^2$

(b). The magnitude of the magnetic torque is $3.20\times10^{35}\ N-m$

Explanation:

Given that,

Mass of neutron $M_{n}= 13M_{s}$

Density of neutron $\rho=4.8\times10^{17}\ kg/m^3$

(a). We need to calculate the rotational inertia

Using formula of rotational inertia for sphere

$I=\dfrac{2}{5}MR^2$ ...(I)

We know that,

$\rho=\dfrac{M}{V}$

Put the value of volume

$\rho=\dfrac{3M_{n}}{4\pi R^3}$

$R^2=(\dfrac{3M_{n}}{4\pi\rho})^{\frac{2}{3}}$

Put the value of R in equation (I)

$I=\dfrac{2}{5}\times M_{n}\times(\dfrac{3M_{n}}{4\pi\rho})^{\frac{2}{3}}$

Put the value into the formula

$I=\dfrac{2}{5}\times(13\times2\times10^{30})^{\frac{5}{3}}\times(\dfrac{3}{4\pi\times(4.8\times10^{17})})^{\frac{2}{3}}$

$I=5.72\times10^{39}\ kg m^2$

The rotational inertia is $5.72\times10^{39}\ kg m^2$ .

(b). We need to calculate the magnitude of the magnetic torque

Using formula of torque

$\tau=I\times \alpha$

Put the value into the formula

$\tau=5.72\times10^{39}\times5.6\times10^{-5}$

$\tau=3.20\times10^{35}\ N-m$

The magnitude of the magnetic torque is $3.20\times10^{35}\ N-m$

Hence, (a). The rotational inertia is $5.72\times10^{39}\ kg m^2$

(b). The magnitude of the magnetic torque is $3.20\times10^{35}\ N-m$

4 0

3 years ago