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madreJ [45]
3 years ago
14

In vertebrates, most of the fluid that ultimately exits the body as urine first enters the nephron tubules by the process of ___

_________.
Physics
1 answer:
Talja [164]3 years ago
7 0
The water and solutes the first enters the nephron tubules undergo a process just before it exits the body as a urine. This process is called filtration. It operates when 20 percent of the solution volume is filtered in the renal corpuscle at a given time. It is reportedly stated that there are around 180 liters of this fluid being filtered by the kidney every day.
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PLEASE HELP, THANK YOU!.. :)
Firlakuza [10]

Answer:

<u><em>First Reaction:</em></u>

^{234}U  =>  ^{230} Th + ^{4}He

<u><em>Second Reaction:</em></u>

^{230} Th => ^{226} Ra + ^{4}He

<u><em>Combined Reaction:</em></u>

^{234} U  => ^{226}Ra + 2( ^{4} He)

8 0
3 years ago
A physical therapy exercise has a person shaking a 5.00 kg weight up and down rapidly. if the barbell is moving at 4.50 m/s, wha
Ivanshal [37]

The magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.

<h3>Magnitude of required force to stop the weight</h3>

The magnitude of the force required to stop the weight in 0.333 seconds is calculated by applying Newton's second law of motion as shown below;

F = ma

F = m(v/t)

F = (mv)/t

F = (5 x 4.5)/0.333

F = 67.6 N

Thus, the magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.

Learn more about force here: brainly.com/question/12970081

#SPJ1

7 0
2 years ago
A car engine changes chemical potential energy into the what energy so the car can move
yuradex [85]
It transforms it to mechanical
7 0
3 years ago
A motorboat accelerates uniformly from a velocity of 6.5 m/s west to a velocity of 1.5 m/s west. If its acceleration was 2.7 m/s
expeople1 [14]

Answer:

<em>The motorboat ends up 7.41 meters to the west of the initial position </em>

Explanation:

<u>Accelerated Motion </u>

The accelerated motion describes a situation where an object changes its velocity over time. If the acceleration is constant, then these formulas apply:

\vec v_f=\vec v_o+\vec a.t

\displaystyle \vec r=\vec v_o.t+\frac{\vec a.t^2}{2}

The problem provides the conditions of the motorboat's motion. The initial velocity is 6.5 m/s west. The final velocity is 1.5 m/s west, and the acceleration is 2.7 m/s^2 to the east. Since all the movement takes place in one dimension, we can ignore the vectorial notation and work with the signs of the variables, according to a defined positive direction. We'll follow the rule that all the directional magnitudes are positive to the east and negative to the west. Rewriting the formulas:

v_f=v_o+a.t

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Solving the first one for t

\displaystyle t=\frac{v_f-v_o}{a}

We have

v_o=-6.5,\ v_f=-1.5,\ a=2.7

Using these values

\displaystyle t=\frac{-1.5+6.5}{2.7}=1.852\ s

We now compute x

\displaystyle x=(-6.5)(1.852)+\frac{(2.7)(1.852)^2}{2}

x=-7,41\ m

The motorboat ends up 7.41 meters to the west of the initial position

5 0
3 years ago
Whats a transverse wave?
Mazyrski [523]

Answer:

a wave vibrating at right angles to the direction of its propagation.

Explanation:

7 0
3 years ago
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