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3 years ago

Describe the structure of an atom

1 answer:

5 0

Answer: Atoms consist of 3 particles: protons, electrons, and neutrons. The nucleus of the atom contains the protons and neutrons. The outer regions contain the electrons

Explanation:

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g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the

MrRa [10]

Answer:

Approximately $-1.58\; \rm rad \cdot s^{-2}$ .

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

$v^2 - u^2 = 2\, a\, x$ ,

where

$v$ is the final velocity of the moving object,
$u$ is the initial velocity of the moving object,
$a$ is the (linear) acceleration of the moving object, and
$x$ is the (linear) displacement of the object while its velocity changed from $u$ to $v$ .

The angular analogue of that equation will be:

$(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ , where

$\omega(\text{final})$ and $\omega(\text{initial})$ are the initial and final angular velocity of the rotating object,
$\alpha$ is the angular acceleration of the moving object, and
$\theta$ is the angular displacement of the object while its angular velocity changed from $\omega(\text{initial})$ to $\omega(\text{final})$ .

For this object:

$\omega(\text{final}) = 0\; \rm rad\cdot s^{-1}$ , whereas
$\omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}$ .

The question is asking for an angular acceleration with the unit $\rm rad \cdot s^{-1}$ . However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

$\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}$ .

Rearrange the equation $(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ and solve for $\alpha$ :

$\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}$ .

7 0

3 years ago

Huck Finn walks at a speed of 0.70 m/s across his raft (that is, he walks perpendicular to the raft’s motion relative to the sho

ZanzabumX [31]

Answer

given,

Hunk Finn speed,v_y= 0.7 m/s

Speed of river,v_x = 1.50 m/s

Assuming the speed of the river is in x-direction.

Speed of Hunk Finn to be in y-direction.

Hunk velocity relative to river =

$x = \sqrt{v_y^2+v_x^2}$

$x = \sqrt{2.74}$

$x =1.66\ m/s$

Speed of Hunk relative to river = 1.66 m/s

direction of the boat

$\theta =tan^{-1}(\dfrac{v_y}{v_x})$

$\theta =tan^{-1}(\dfrac{0.7}{1.50})$

$\theta = 25.01^{\circ}$

Hence, angle of the raft is 24.01°

6 0

4 years ago

Two trains approach each other on parallel tracks. Each has a speed of 95 kph relative to the ground. If they are initially 8.5

aliya0001 [1]

Answer: 2.83 minutes

Explanation:

It is understood that trains are approaching. That is, they have speeds of equal magnitude but opposite. When train A travels x meters northbound, then train B travels the same distance southbound.

Therefore trains approach at a speed of:

$V = 95 +95 = 190\ km / h$

Then:

$V = \frac{x}{t}\\\\t = \frac{x}{V}$

Where x is the distance between the trains

$x = 8.5\ km$

So the time in which both trains meet is:

$t =\frac{8.5\ km}{180\ \frac{km}{h}}\\\\t=0.04722\ hours$

This is:

$0.04722*\frac{60\ minutes}{1\ hour} = 2.83\ minutes$

<em />

<em>How long will it be before they reach one another ?</em>

<h3>2.83 minutes</h3>

3 0

3 years ago

A cricketer moves his hands backwards when holding a catch

lisov135 [29]

By doing this he increases the time due to which the rate of change momnetemdecreases. that means the ball comes to rest gently and does not hurt the fielder. but, if does not move his hand backwards, then the ball would come to rest in a fraction of second with a high rate of change of momentum. the ball will exert a lot of force on the hands of the fielder which will hurt him.

3 0

3 years ago

Question 2 (1 point)

love history [14]

Answer:

neutral

Explanation:

3p - 3e = 0 and that leaves 2 neutrons so it will be neutral

6 0

4 years ago

Read 2 more answers