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2 years ago

7

The ph balance is maintained by which organ?

1 answer:

Lena [83]2 years ago

7 0

The Kidneys And Lungs.

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Define malleability?<br>

ad-work [718]

Answer:

Explanation:

the quality or state of being malleable: such as. a : capability of being shaped or extended by hammering, forging, etc. the malleability of tin.

5 0

3 years ago

How do you find the range of a data set?

Helga [31]

The answer is A.

The answer is A.

The answer is A.

6 0

3 years ago

The pressure in a traveling sound wave is given by the equation ΔP = (1.78 Pa) sin [ (0.888 m-1)x - (500 s-1)t] Find (a) the pre

frozen [14]

Answer:

a) $P_m=1.78\ Pa$

b) $f=79.5775\ Hz$

c) $\lambda=7.076\ m$

d) $v=563.06\ m.s^{-1}$

Explanation:

<u>Given equation of pressure variation:</u>

$\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]$

We have the standard equation of periodic oscillations:

$\Delta P=P_m\ sin\ (kx-\omega.t)$

<em>By comparing, we deduce:</em>

(a)

amplitude:

$P_m=1.78\ Pa$

(b)

angular frequency:

$\omega=2\pi.f$

$2\pi.f=500$

∴Frequency of oscillations:

$f=\frac{500}{2\pi}$

$f=79.5775\ Hz$

(c)

wavelength is given by:

$\lambda=\frac{2\pi}{k}$

$\lambda=\frac{2\pi}{0.888}$

$\lambda=7.076\ m$

(d)

Speed of the wave is gives by:

$v=\frac{\omega}{k}$

$v=\frac{500}{0.888}$

$v=563.06\ m.s^{-1}$

8 0

2 years ago

A machine has a mechanical advantage of 4.5. What force is put out by the machine if the force applied to the machine is 800 N?

Ghella [55]

If the machine's mechanical advantage is 4.5, that means that

Output force = (4.5) x (Input force) .

We know the input force, and we need to find the output force. Rather than wander around the room looking at the floor while our hair smolders, let's try putting the numbers we know into the equation I wrote up there. OK ?

Output force = (4.5) x (Input force)

Output force = (4.5) x (800 N)

Now dooda multiplication:

<em>Output force = 3,600 N</em> .

That's exactly what the question asked for. So we're done !

3 0

2 years ago

A 75 kg baseball player runs at a velocity of 6 m/s before sliding to a stop at second base. a. What is the kinetic energy of th

lana [24]

Answer:

a. $\displaystyle k_o=1350\ J$

b. $\displaystyle k_1=0\ J$

c. $\Delta k=-1350\ J$

d. $W=-1350\ J$

e. $F=-675\ N$

Explanation:

<u>Work and Kinetic Energy </u>

When an object moves at a certain velocity v0 and changes it to v1, a change in its kinetic energy is achieved:

$\Delta k=k_1-k_0$

Knowing that

$\displaystyle k=\frac{mv^2}{2}$

We have

$\displaystyle \Delta k=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}$

The work done by the force who caused the change of velocity (acceleration) is

$\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}$

If we know the distance x traveled by the object, the work can also be calculated by

$W=F.x$

Being F the force responsible for the change of velocity

The 75 kg baseball player has an initial velocity of 6 m/s, then he slides and stops

a. Before the slide, his initial kinetic energy is

$\displaystyle k_o=\frac{mv_0^2}{2}$

$\displaystyle k_o=\frac{(75)6^2}{2}$

$\boxed{\displaystyle k_o=1350\ J}$

b. Once he reaches the base, the player is at rest, thus his final kinetic energy is

$\displaystyle k_1=\frac{(75)0^2}{2}$

$\boxed{\displaystyle k_1=0\ J}$

c. The change of kinetic energy is

$\Delta k=k_1-k_0=0\ J-1350\ J$

$\boxed{\Delta k=-1350\ J}$

d. The work done by friction to stop the player is

$W=\Delta k=k_1-k_0$

$\boxed{W=-1350\ J}$

e. We compute the force of friction by using

$W=F.x$

and solving for x

$\displaystyle F=\frac{W}{x}$

$\displaystyle F=\frac{-1350\ J}{2\ m}$

$\boxed{F=-675\ N}$

The negative sign indicates the force is against movement

6 0

2 years ago