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4 years ago

During which stage of prenatal development does breathing movement begin?

Physics

2 answers:

Damm [24]4 years ago

5 0

These begin at 16 weeks of pregnancy

Juliette [100K]4 years ago

5 0

Answer:

Because it's asking stage not time period the answer is Fetal

If this helped, please help me with a like ;)

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A girl drops a pebble from a high cliff into a lake far below. She sees the splash of the pebble hitting the water 2.00s later.

Natali5045456 [20]

Answer:

19.62 ms

Explanation:

t = Time taken = 2 s

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (downward direction is taken as positive)

Equation of motion

$v=u+at\\\Rightarrow v=0+9.81\times 2\\\Rightarrow v=19.62\ m/s$

The speed of the pebble when it hit the water is 19.62 ms

3 0

3 years ago

How does the total momentum of the docked vehicle and station compare to the momentum of each object before the

Xelga [282]

The total momentum before the docking maneuver is and after the docking maneuver is

Explanation:

Linear momentum (generally just called momentum) is defined as mass in motion and is given by the following equation:

Where is the mass of the object and its velocity.

According to the conservation of momentum law:

"If two objects or bodies are in a closed system and both collide, the total momentum of these two objects before the collision will be the same as the total momentum of these same two objects after the collision ".

This means, that although the momentum of each object may change after the collision, the total momentum of the system does not change.

Now, the docking of a space vehicle with the space station is an inelastic collision, which means both objects remain together after the collision.

Hence, the initial momentum is:

Where:

is the mass of the vehicle

is the velocity of th vehicle

is the mass of the space station

is the velocity of the space station

And the final momentum is:

Where:

is the velocity of the vehicle and space station docked

5 0

4 years ago

Consider two massless springs connected in parallel. Springs 1 and 2 have spring constants k1 and k2 and are connected via a thi

77julia77 [94]

Answer:

k1 + k2

Explanation:

Spring 1 has spring constant k1

Spring 2 has spring constant k2

After being applied by the same force, it is clearly mentioned that spring are extended by the same amount i.e. extension of spring 1 is equal to extension of spring 2.

x1 = x2

Since the force exerted to each spring might be different, let's assume F1 for spring 1 and F2 for spring 2. Hence the equations of spring constant for both springs are

k1 = F1/x -> F1 =k1*x

k2 = F2/x -> F2 =k2*x

While F = F1 + F2

Substitute equation of F1 and F2 into the equation of sum of forces

F = F1 + F2

F = k1*x + k2*x

= x(k1 + k2)

Note that this is applicable because both spring have the same extension of x (I repeat, EXTENTION, not length of the spring)

Considering the general equation of spring forces (Hooke's Law) F = kx,

The effective spring constant for the system is k1 + k2

3 0

3 years ago

The type of error which is corrective

vodka [1.7K]

Answer:

ummm i dont know what...

4 0

3 years ago

The acceleration of an object would increase if there was an increase in the

xxMikexx [17]

As per Newton's II law we know that

$F_{net} = ma$

here we know that

$F_{net} = F_{ap} - F_f$

so here we will have

$a = \frac{F_{ap} - F_f}{m}$

so here if we need to increase the acceleration we need to increase the applied force while on increasing the mass or on increasing the friction force the acceleration will decrease.

So here correct answer will be

<em>A) force on the object.</em>

4 0

4 years ago