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3 years ago

How is urine produced

Physics

2 answers:

Tom [10]3 years ago

5 0

Urine is a liquid by-product of metabolism in humans and in many other animals. Urine flows from the kidneys through the ureters to the urinary bladder. Urination results in urine being excreted from the body through the urethra.

Citrus2011 [14]3 years ago

3 0

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A 290 gg bird flying along at 6.2 m/sm/s sees a 9.0 gg insect heading straight toward it with a speed of 34 m/sm/s (as measured

Murrr4er [49]

Answer:

The bird's speed immediately after swallowing is 4.98 m/s.

Explanation:

Given that,

Mass of bird = 290 g

Speed = 6.2 m/s

Mass of sees = 9.0 g

Speed = 34 m/s

We need to calculate the bird's speed immediately after swallowing

Using conservation of momentum

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{3}$

Put the value into the formula

$0.290\times6.2+0.009\times(-34)=(0.290+0.009)\times v_{3}$

$v_{3}=\dfrac{0.290\times6.2-0.009\times34}{(0.290+0.009)}$

$v_{3}=4.98\ m/s$

Hence, The bird's speed immediately after swallowing is 4.98 m/s.

6 0

3 years ago

Because quantum mechanics is physics that describes the interactions of very small objects (i.e. molecules, atoms, and electrons

stira [4]

Ha! Lot of words but the question itself is easy.
The answer is 2.5 times 10 to the 5th power.
The main part of the numbers has the decimal point placed after the first digit.
Then for what number of power, you just count the number of decimal places moved.
I hope this helps you.

5 0

3 years ago

Read 2 more answers

An object moving due to gravity can be described by the motion equation y=y0+v0t−12gt2, where t is time, y is the height at tha

LenaWriter [7]

Answer:

$t=6.4534 s$

Explanation:

This is an exercise where you need to use the concepts of free fall objects

Our knowable variables are initial high, initial velocity and the acceleration due to gravity:

$y_{0}=75m$

$v_{oy} =20m/s$

$g=9.8 m/s^{2}$

At the end of the motion, the rock hits the ground making the final high y=0m

$y=y_{o}+v_{oy}*t-\frac{1}{2}gt^{2}$

If we evaluate the equation:

$0=75m+(20m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$

This is a classic form of Quadratic Formula, we can solve it using:

$t=\frac{-b ± \sqrt{b^{2}-4ac } }{2a}$

$a=-4.9\\b=20\\c=75$

$t=\frac{-(20) + \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=-2.37s$

$t=\frac{-(20) - \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=6.4534s$

Since the time can not be negative, the reasonable answer is

$t=6.4534s$

8 0

3 years ago

Two workers pull horizontally on a heavy box but one pulls twice as hard as the other. The larger pull is directed at 25.0 west

icang [17]

1) Call F1 the larger force and F1x and F1y its its x-and-y- components.respectively.

I will use the complementary angle: 90 - 25 = 65 to work with the normal convention.

=> cos(65) = F1x / F1 => F1x = - F1*cos(65) (I choose negative as the west direction)

=> sin(65) = F1y / F1 => F1y = F1*sin(65) (I choose positive the north direction)

2) Call F2 the shorter force and F2x and F2y its components

=> cos(x) = F2x / F2 => F2x = F2*cos(x)

=> sin(x) = F2y / F2=> F2y = F2*sin(x)

3) You know that:

- F1 = 2F2
- The net force in the y direction is 430 N
- The net force in the x direction is 0

a) F1x + F2x = 0

=> -F1*cos(65) + F2*sin(x) = 0

=> F1*cos(65) = F2 sin(x) => sin(x) = [F1/F2] cos(65)

Remember F1 = 2F2 => F1/F2 = 2 => sin(x) = 2 cos(65) = 0.84524

=> x = arcsin(0.84524) = 57.7

b) F1y + F2y = 430 =>

F1 sin(65) + F2*sin(57.7) = 430 =>

0.9060F1 + 0.84524F2 430

F1 = 2F2 => 0.9060*2F2 + 0.84524F2 = 430 => 1.7512F2 = 430

=> F2 = 430 / 1.7512 = 245.54 N

=> F1 = 2*245.54 =491.1N

There you have the two forces.

The angle of the shorter force is 57.7 measured from the east to the north (this is north of east), which would be 90 - 57.7 = 32.3 degrees east of north..

Then the shorter force is 245.5 N at 32.3 degrees east of north

And the larger force is 491.1 N at 25.0 degrees west of north.

3 0

4 years ago

What is Hooke's law? what is meant by elastic limit? please answer me

Nutka1998 [239]

Answer:

Explanation:

The law is about how object will be deformed when certain amount force is applied. Elastic limit is a point where object will not return its previous shape after being deformed

6 0

3 years ago