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3 years ago

What functions normally performed by the placenta would the hatchery have to perform so that the embryos survive

1 answer:

6 0

The functions that would be performed both by the placenta and the hatchery so that the embryos will survive is to maintain the temperature of the embryos. The temperature should also be at the temperature where the embryos would thrive and develop.

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An object is placed near a concave mirror having a radius of curvature of magnitude 60 cm. How far should you place the object f

Viefleur [7K]

Answer:

u = 18 cm

Explanation:

given,

radius of curvature = 60 cm

magnification of mirror = 2.5

distance of object = ?

R = 2 f

f = R/2

f = 60/2 = 30 cm

$m = -\dfrac{v}{u}$

$2.5 = -\dfrac{v}{u}$

v = -2.5 u

now,

Using mirror formula

$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

$\dfrac{1}{30} = \dfrac{1}{-2.5u} + \dfrac{1}{u}$

$\dfrac{1}{30} = \dfrac{0.6}{u}$

u = 0.6 x 30

u = 18 cm

distance of object be equal to u = 18 cm

6 0

3 years ago

The volume in the pump when the pump piston is all the way down represents the _______

sveticcg [70]

The volume in the pump when the pump piston is all the way down represents the end systolic volume.

<h3>What is systolic volume?</h3>

End systolic volume is the volume of blood in a ventricle at the end of contraction, or systole, and the beginning of filling, or diastole.

End systolic volume is the lowest volume of blood in the ventricle at any point in the cardiac cycle.

Thus, the volume in the pump when the pump piston is all the way down represents the end systolic volume.

Learn more about volume here: brainly.com/question/1972490

#SPJ12

7 0

2 years ago

Calculate the force necessary to accelerate a 10 kg table from<br> O m/s to 4 m/s in 2 seconds.

yanalaym [24]

Answer:

a= v/t

a = 4/2

a = 2 m/s^2

And F = M a

F = 10 × 2

F = 20 N

6 0

3 years ago

In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.

Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement of the object

t is the time elapsed

$\omega$ is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

$\theta=2\pi$ rad

And the time taken is

$t=95 min \cdot 60 =5700 s$

Therefore, the angular velocity of the telescope is

$\omega=\frac{2\pi}{5700}=0.0011 rad/s$

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

$v=\omega r$

where

v is the linear velocity

$\omega$ is the angular velocity

r is the radius of the circular orbit

In this problem:

$\omega=0.0011 rad/s$ is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

$r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m$

Therefore, the linear velocity of the telescope is:

$v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s$

4 0

3 years ago

The ideal mechanical advantage of an inclined plane is 3.5, and its efficiency is 0.6. What is the mechanical advantage of the i

Nadya [2.5K]

Mechanical advantage = ideal mechanical advantage x efficiency = 3.5 x 0.6 = 2.1
The mechanical advantage of the inclined plane is 2.1

7 0

3 years ago