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3 years ago

13

List the several mechanisms you have explored that change the glomerular filtration rate. how does each mechanism specifically a

lter the glomerular filtration rate?

1 answer:

stiks02 [169]3 years ago

3 0

The formal operational stage starts at around age twelve and keeps going into adulthood. Amid this time, individuals build up the capacity to consider conceptual ideas. Abilities, for example, intelligent thought, deductive thinking, and precise arranging likewise rise amid this stage.

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The area of the pond is approximately equal to the area of a circle with radius 297m. Find the mass of the ice. Answer in kilogr

True [87]

Answer:

<em>mass of the ice is 254980463.8T kg</em>

<em>where T is the value of the thickness omitted in the question.</em>

Explanation:

The ice on Walden Pond is .......... thick. The area of the pond is approximately equal to the area of a circle with radius 297 m. Find the mass of the ice. Answer in kg.

<em>The value of the thickness of the ice T is omitted, but I will show the solution, and the real answer can be gotten by multiplying the final calculated answer here by the thickness of the ice omitted.</em>

Given the radius of the equivalent circle of the ice = 297 m'

the area of the ice can be gotten from area A = $\pi r^{2}$ = $3.142*297^{2}$ = 277152.678 m^2

recall that the density of ice p ≅ 920 kg/m^3

also,

density of ice p = (mass of ice, m) ÷ (volume of ice, v)

i.e p = m/v

and,

m = pv

substituting the value of the density of water p into the equation, we have,

mass of the ice, m = 920v ....... equ 1

The volume of the ice above will be = (area of the ice, A) x (thickness of the ice, T)

i.e v = AT

substituting the value of area A into the equation, we have

v = 277152.678T ......equ 2

substitute value of v into equ 1

mass of the ice, m = 920 x (277152.678T)

mass of the ice, m = 254980463.8T kg

where T is the thickness of the ice

NB: To get the mass, multiply this answer with the thickness T given in the question.

7 0

3 years ago

A backpack has a mass of 8 kg. It is lifted and given 54.9 J of gravitational potential energy. How high is it lifted? Accelerat

sweet [91]

Potencial Energy=hma
Where
Potencial Energy =E= 54.9J
h=?
m=8kg
a=9.8m/s^2
You need to know that 1 J=1(kgm^2)/s^2

Isolate h=E/(ma)
h=(54.9)/(8*9.8)

8 0

3 years ago

Read 2 more answers

Calculate the pressure exerted on the floor when an

skad [1K]

P = F / A

P = 2400 / 4

P = 24 × 100 / 4

P = 6 × 100

P = 600 N/m^2

7 0

3 years ago

You are a member of a geological team in Central Africa. Your team comes upon a wide river that is flowing east. You must determ

Dovator [93]

Answer:

(a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

Explanation:

Given that,

Constant speed = 6.00 m/s

Time = 20.1 sec

Speed = 9.00 m/s

Time = 11.2 sec

We need to write a equation for to travel due north across the river,

Using equation for north

$v^2-c^2=\dfrac{w^2}{t^2}$

Put the value in the equation

$6.00^2-c^2=\dfrac{w^2}{(20.1)^2}$

$36-c^2=\dfrac{w^2}{404.01}$ ....(I)

We need to write a equation for to travel due south across the river,

Using equation for south

$v^2-c^2=\dfrac{w^2}{t^2}$

Put the value in the equation

$9.00^2-c^2=\dfrac{w^2}{(11.2)^2}$

$81-c^2=\dfrac{w^2}{125.44}$ ....(II)

(a). We need to calculate the wide of the river

Using equation (I) and (II)

$45=\dfrac{w^2}{125.44}-\dfrac{w^2}{404.01}$

$45=w^2(0.00549)$

$w^2=\dfrac{45}{0.00549}$

$w=\sqrt{\dfrac{45}{0.00549}}$

$w=90.5$

We need to calculate the current speed

Using equation (I)

$36-c^2=\dfrac{(90.5)^2}{(20.1)^2}$

$36-c^2=20.27$

$c^2=20.27-36$

$c=\sqrt{15.73}$

$c=3.96\ m/s$

(b). We need to calculate the shortest time

Using formula of time

$t=\dfrac{d}{v}$

$t=\dfrac{90.5}{6}$

$t=15.0\ sec$

We need to calculate the distance

Using formula of distance

$d=vt$

$d=3.96\times15.0$

$d=59.4\ m$

Hence, (a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

3 0

4 years ago

Protons are located Select one:

asambeis [7]

Option C: in the nucleus with neutrons.

5 0

3 years ago

Read 2 more answers