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3 years ago

When Coach Kwan notices that a player is getting tired, she takes out the tired player and substitutes a fresh player.

Physics

2 answers:

drek231 [11]3 years ago

8 0

Answer:

is replacement

Explanation:

Wittaler [7]3 years ago

5 0

The answer is a

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A red blood cell contains 4.8 107 free electrons. What is the total charge of these electrons in the red blood cell?

tatuchka [14]

Answer:

Charge, $q=7.68\times 10^{-12}\ C$

Explanation:

It is given that,

The number of electron in a RBCs, $n=4.8\times 10^7$

We need to find the total charge of these electrons in the red blood cell. Let it is q. Using the quantization of charge as follows :

q = ne

e is the change on electron

$q=4.8\times 10^7\times 1.6\times 10^{-19}\\\\q=7.68\times 10^{-12}\ C$

So, the net charge is $7.68\times 10^{-12}\ C$ .

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3 years ago

A German scientist, George Ohm, discovered the relationship between current, voltage, and resistance. This became known as Ohm's

Korvikt [17]

Choice A ... I = V/R ... is a correct form of Ohm's Law.

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3 years ago

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Is water still part of the cycle when it is underground? Explain.

Marysya12 [62]

Yes it is, even though the water is underground, it can still be evaporated. Large amounts of water are stored in the ground. The water is still moving, possibly very slowly, and it is still part of the water cycle. Most of the water in the ground comes from precipitation that infiltrates downward from the land surface.

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3 years ago

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if Luke pushes with 22 N of force on the wagon that is now filled with bricks but he cannot make you move how much work did Luke

madam [21]

Answer: He did none.

Explanation:

Even though he may have felt like he did work, none was done.

The only way for work to happen is if the object moves.

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3 years ago

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Two small particles of mass m1 and mass m2 attract each other with a force that varies inversely with the cube of their separati

Naya [18.7K]

Answer:

$r_{cm}\ =\ \dfrac{m_2d\ +\ m_1v_0 (t_1\ -\ t_0)}{m_1\ +\ m_2}$

Explanation:

Given,

mass of the first particle = $m_1$
velocity of the first particle = $v_o$
mass of the second particle = $m_2$
velocity of the second particle = $v_2 = 0$
Time interval = $(t_1\ -\ t_o)$

Let $v_{cm}$ be the initial velocity of the center of mass of the system of particle at time $t_o$

$\therefore v_{cm}\ =\ \dfrac{m_1v_1\ +\ m_2v_2}{m_1\ +\ m_2}\\\Rightarrow v_{cm}\ =\ \dfrac{m_1v_0}{m_1\ +\ m_2}$

Assuming that the first particle is at origin, distance of the second particle from the origin is 'd'

$x_1\ =\ 0$
$x_2\ =\ d$

Center of mass of the system of particles

$x_{cm}\ =\ \dfrac{m_1x_1\ +\ m_2x_2}{m_1\ +\ m_2}\\\Rightarrow x_{cm}\ =\ \dfrac{m_2d}{m_1\ +\ m_2}\\$

Hence, at time $t_0$ , the center of mass of the system is at $x_0\ =\ \dfrac{m_2d}{m_1\ +\ m_2}$ at an initial speed of $v_{cm}$

Both the particles are assumed to be the point masses, therefore at the time $t_1$ the center of mass is at the position of the second particle which should be equal to the total distance traveled by the first particle because the second particle is at rest.

Let $r_{cm}$ be the distance traveled by the center of mass of the system of particles in the time interval $(t_1\ +\ t_0)$

From the kinematics,

$s\ =\ x_0\ +\ vt\\\Rightarrow r_{cm}\ =\ x_{cm}\ +\ v_{cm}{t_1\ -\ t_0}\\\Rightarrow r_{cm}\ =\ \dfrac{m_2d}{m_1\ +\ m_2}\ +\ \left ( \dfrac{m_1v_0}{m_1\ +\ m_2}\ \right )\times (t_1\ -\ t_0)\\\Rightarow r_{cm}\ =\ \dfrac{m_2d\ +\ m_1v_0 (t_1\ -\ t_0)}{m_1\ +\ m_2}$

Hence, this is the required distance traveled by the first mass to collide with the second mass which is at rest.

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3 years ago