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Rudiy27
3 years ago
13

Which type of tissue is most likely to prevent a kidney from "floating"?

Physics
1 answer:
Sedaia [141]3 years ago
6 0

c. Adipose tissue

Explanation:

The type of tissue that most likely will prevent a kidney from floating is an adipose tissue.

  • Kidney floating is a rare condition in which the kidney is out of position.
  • It is a rare condition in which the kidney drops to the pelvis when they stand up.
  • The condition is medically known as nephroptosis.
  • It is usually caused when the ligaments attaching the kidney to the body weakens.
  • Each kidney is held in place by the renia fascia and is surrounded by the adipose tissues.

Learn more:

Kidney brainly.com/question/9111691

#learnwithBrainly

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Answer:

D is the answer

Explanation:

But do u know if im right test it out then surprise

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2 years ago
A person pushes on a box with a force of 500N. If the box does not move, what is the force of static friction on the box?
Alexxx [7]

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c

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4 0
3 years ago
Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after
adoni [48]

Answer: v = 1.19 * 10^{6} m/s

Explanation: q = magnitude of electronic charge = 1.609 * 10^{-19} c

mass of an electronic charge = 9.10 * 10^{-31} kg

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = \frac{mv^{2} }{2},  potential energy = qV

hence, \frac{mv^{2} }{2} = qV

\frac{9.10 *10^{-31} * v^{2}  }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31}  * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s

7 0
2 years ago
While driving on a rural road, your right wheels run off the pavement. You should hold the steering wheel firmly and
Vsevolod [243]

Answer:

The answer is C. Steer in a straight line while gently slowing down

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The following are advised when your cars go off the pavement while driving;

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ensure you hold on to your steering wheel tightly.

keep Steering straight ahead.

ensure you Stay on the shoulder.

Ease up on the accelerator and brake gently.

When you  know you can safely do so, turn back on the road at a much lower speed.

3 0
2 years ago
An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu
ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, qE \times S + mg \times S = 0.5 \times mv^{2}

     1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight  and the above equation will be as follows.

   (1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
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         v = sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}

           = 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0
2 years ago
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