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1 year ago

What is the speed of each neutron as they crash together? keep in mind that both neutrons are moving?

Physics

2 answers:

Oksanka [162]1 year ago

8 0

Answer:

The speed, as they crash together, is the same than before the crash, but with opposite directions. In this context, we recur to the collision theory in physics, which says that the amount of movement of each particle before the collision is the same than after the collision, due to the conservation of the amount of movement, which can be described like this: $\Delta p=0$ . This expression means that in a isolated environment (which is present in a neutron's collision), the total amount of movement doesn't change.

Therefore, <em>the speed of neutrons is the same before and after the collision</em>. Specifically the number of this speed depends on the experiment.

Sophie [7]1 year ago

7 0

This involves shooting electrons (from an accelerator) at a target or protons. This technique provided evidence for the existence of quarks. <span>proton-antiproton scattering as well.
</span>hope this helps

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A ball is dropped from the rest from a height of 6. 0 meters above the ground. The ball falls freely and reaches the ground at t

igor_vitrenko [27]

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2 years ago

Read 2 more answers

Emmy kicks a soccer ball up at an angle of 45° over a level field. She watches the ball's trajectory and notices that it lands,

Elenna [48]

Let $u$ be the initial velocity of the soccer ball at an angle of inclination of $\theta_0$ with the positive x-axis.

Given that:

$\theta_0=45^{\circ}$

The horizontal distance covered by the projectile=20 m

Time of flight, $t_f=2$ seconds

Acceleration due to gravity, $g= 10 m/s^2$ downward.

As "north" and "up" as the positive x ‑ and y ‑directions, respectively.

So, $g= -10 m/s^2$

As the acceleration due to gravity is in the vertical direction, so the horizontal component of the initial velocity remains unchanged.

The x-component of the initial velocity, $u_x=u\cos\theta_0$ .

The horizontal distance covered by the projectile $= u_x\times t_f$

$\Rightarrow u_x\times t_f=20$

$\Rightarrow u_x\times 2=20$

$\Rightarrow u_x=10$ m/s

So, the horizontal component of the velocity is 10 m/s which is constant and the graph has been shown in the figure (i).

Now, $u\cos(45^{\circ})=10$ [as $u_x=u\cos\theta_0$ ]

$\Rightarrow u=10\sqrt{2}$ m/s.

The vertical component of the initial velocity,

$u_y= u\sin\theta_0$

$\Rightarrow u_y=10\sqrt{2}\sin(45^{\circ})$

$\Rightarrow u_y=10$ m/s

Let v be the vertical component of the velocity at any time instant t.

From the equation of motion,

$v=u+at$

where u: initial velocity, v: final velocity, a: constant acceleration, and t: time taken to change the velocity from u to v.

In this case, we have $u=u_y, a= -10 m/s^2$ .

So at any time instant, t.

$v=u_y+(-10)t$

$\Rightarrow v=10-10t$

The vertical component of the velocity, v, is the function of time and related as $v=10-10t$ .

This is a linear equation.

At 2 second, the vertical component of the velocity

v=10-10x2=-10 m/s.

The graph has been shown in figure (ii).

7 0

2 years ago

One end of a piano wire is wrapped around a cylindrical tuning peg and the other end is fixed in place. The tuning peg is turned

liq [111]

Answer:

$T = 3183 N$

Explanation:

When the screw is turned by two turns then change in the length of the wire is given as

$\Delta L = 2(2\pi r)$

$\Delta L = 4\pi r$

$\Delta L = 4(\pi)(1.0 mm)$

$\Delta L = 12.56 mm$

now we know by the formula of Young's modulus

$Y = \frac{T/A}{\Delta L/L}$

so we have

$T = \frac{AY \Delta L}{L}$

$T = \frac{\pi (0.55 \times 10^{-3})^2(2.0 \times 10^{11})(12.56 \times 10^{-3})}{0.75}$

$T = 3183 N$

3 0

2 years ago

Does the rankine degree represent a larger or smaller temperature unit than the kelvin degree

zvonat [6]

The best and most correct answer among the choices provided by the question is the first choice, larger.

Rankine is Fahrenheit + 460 , while Kelvin is Celsius + 273. We all know that Fahrenheit has larger number compared to kelvin , thus rankine is much larger.
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.

8 0

2 years ago

two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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2 years ago