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Vladimir [108]
3 years ago
9

3. Which of the following is NOT a function of the liver?

Physics
2 answers:
Inga [223]3 years ago
8 0

Answer:

C) to produce white blood cells

Explanation:

Your liver does all of the others things except produce white blood cells

If it helps consider making brianliest (it helps alot)

mash [69]3 years ago
7 0
I think it’s C, that persons answer is good
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A positively charged light metal ball is suspended between two oppositely charged metal plates on
Ipatiy [6.2K]

Answer:

The positively charged ball moves between both charged plates till the plates and the ball all become neutral.

Check Explanation for more.

Explanation:

Let the ball be in square brackets, and the plates in normal brackets.

(+) [+] (-)

From the law that like charges repel and unlike charges attract.

The positive ball would go first to the negatively charged plate. After which, the ball would hold more negative charges overall than before.

Because the ball is now more negatively charged, it then travels towards the positive plate. In the same manner, the ball would transfer negative electrons to the positive plate.

So, when leaving the positive plate, the ball would be more positive and be drawn towards the negative plate once more. In doing so, it would make the negative plate more positive.

Then, the ball again holds more negative electrons and is drawn towards the positive plate once more.

This back and forth process continues until the once-positive and once-negative plates become neutral, that is, they are discharged.

The ball hanging on the insulated thread becomes neutral too at this point.

Hope this Helps!!!

8 0
3 years ago
A sound wave enters a new medium where sound travels faster. How does this affect the frequency and wavelength of the sound?
iren [92.7K]
The frequency doesn't change. If the wavespeed increases, then the wavelength must also increase ... It's just the distance the wave travels during each complete wiggle.
6 0
3 years ago
Read 2 more answers
Find the ratio of the diameter of iron to copper wire, if they have the same resistance per unit length (as they might in househ
Natasha_Volkova [10]

Answer:

The ratio of the diameter of iron to Cu is;

\frac{d{Fe}   }{ d{Cu}   } =\sqrt{\frac{p_{Fe} }{ p_{Cu} }}

Explanation:

R=(ρL)/A

  • R is resistance,
  • L is length,
  • A is area,
  • ρ is resistivity
  • d is diameter

from the question the two materials have the same resistance per unit length.

\frac{R}{L}= \frac{p}{A}

\frac{R}{L}   for iron = \frac{R}{L}  for copper

This means we can equate ρ/A for both materials.

\frac{p_{Fe} }{A_{Fe} } =\frac{p_{Cu} }{A_{Cu} }

re-arranging the equation we have,

\frac{A_{Fe}}{A_{Cu} } =\frac{p_{Fe} }{ p_{Cu} }

A=\pi \frac{d^{2} }{4}

\frac{A_{Fe}}{A_{Cu} } =\frac{d^{2}{Fe}   }{ d^{2}{Cu}   }

\frac{d^{2}{Fe}   }{ d^{2}{Cu}   } =\frac{p_{Fe} }{ p_{Cu} }

\frac{d{Fe}   }{ d{Cu}   } =\sqrt{\frac{p_{Fe} }{ p_{Cu} }}

5 0
3 years ago
a driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di
daser333 [38]

Answer:

v = 98.75 km/h

Explanation:

Given,

The distance driver travels towards the east, d₁ = 135 km

The time period of the travel, t₁ = 1.5 h

The halting time, tₓ = 46 minutes

The distance driver travels towards the east, d₂ = 215 km

The time period of the travel, t₁ = 2 h

The average speed of the vehicle before stopping

                                    v₁ = d₁/t₁

                                        = 135/1.5

                                       = 90 km/h

The average speed of vehicle after stopping

                                    v₂ = d₂/t₂

                                         = 215/2

                                        = 107.5 km/h

The total average velocity of the driver

                                      v = (v₁ +v₂) /2

                                         = (90 + 107.5)/2

                                         = 98.75 km/h

Hence, the average velocity of the driver, v = 98.75 km/h

7 0
4 years ago
What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate
Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The  values is  E =248.2 \  N/C

Explanation:

From the question we are told that

   The  diameter is  d =  12 \  cm  =  0.12 \ m

    The charge  is  Q =  4.4 nC  =  4.4 *10^{-9} \  C

    The  distance from the center  is  k =  0.1 \ mm   =  1*10^{-4} \  m

Generally the radius is mathematically represented as

        r =  \frac{d}{2}

=>     r =  \frac{0.12}{2}

=>       r =  0.06 \  m

Generally electric field is mathematically represented as  

       E =  \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 +  k^2 } } ]

substituting values  

      E =  \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 +  (1.0*10^{-4})^2 } } ]

     E =248.2 \  N/C

4 0
3 years ago
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