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3 years ago

Compare and contrast the processes of fission and fusion

2 answers:

6 0

Atomic nuclei lighter than iron and nickel, energy can be extracted by combining iron and nickel nuclei together through nuclear fusion. For atomic nuclei heavier than iron or nickel, energy can be released by splitting the heavy nuclei through nuclear fission.

cricket20 [7]3 years ago

3 0

Fusion is joining two things together while Fission is Dividing two things into pieces

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Hằng số phổ biến chất khí

padilas [110]

Answer:

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Explanation:

8 0

2 years ago

. Write the following in the scientific notation of 0.0000002400m

aleksandr82 [10.1K]

Explanation:

<h2>Steps :</h2>

Move decimal from left to right =0 000000240.0
Then count the numbers before decimal and write it like this =240.0x10 power-9
That's all

hope it help

7 0

2 years ago

In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.

HACTEHA [7]

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is $41.9^{\circ}$

Explanation:

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

$s=u_y t + \frac{1}{2}at^2$ (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

$u_y = u sin \theta$ is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

$\theta=40.0^{\circ}$ is the angle of projection

$u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s$

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (downward)

Substituting the numbers, we get

$-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0$

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

$d=u_x t$

where

$u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s$ is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

$d=(9.19)(1.778)=16.34 m$

In this second case,

$\theta=42.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.1t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s$

So, the range of the shot is

$d=u_x t = (8.85)(1.851)=16.38 m$

In this third case,

$\theta=45^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.5t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s$

So, the range of the shot is

$d=u_x t = (8.49)(1.925)=16.34$ m

In this 4th case,

$\theta=47.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.8t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s$

So, the range of the shot is

$d=u_x t = (8.11)(1.981)=16.07 m$

From the previous parts, we see that the maximum range is obtained when the angle of releases is $\theta=42.5^{\circ}$ .

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

$s-u sin \theta t + \frac{1}{2}gt^2=0$

The solutions of this quadratic equation are

$t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}$

This is the time of flight: so, the horizontal range is

$d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta$

It can be found that the maximum of this function is obtained when the angle is

$\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})$

Therefore in this problem, the angle which leads to the maximum range is

$\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0

3 years ago

A tennis ball is tossed up off a building with a velocity of 22m/s. It takes 6.4s to reach the ground. How high is the building

Amanda [17]

The height of the tennis ball,relative to the ground is H=h max+h-->h max-the maximum height that the tennis ball reaches relative to the roof of the building; h-the height of the building;h max =v0^2/2g=24,2m(g=10m/s^2).H=gt^2/2=>24,2+h=gt^2/2=>h=gt^2/2-24,2=180,6m

4 0

3 years ago

What is the difference between a negative feedback system and a positive feedback system?

Anni [7]

Answer:

Explanation:

The negative feedback is responsible for maintaining equilibrium (stability) in a system as it lessens effects, which is contrary to positive feedback which increases reaction and moves a system further away from equilibrium (stability), It does so by amplifying the effects of a product or event and occurs when something needs to happen quickly. e.g

Insulin lowers down blood sugar levels, so when the body detects that it has too much sugar, the pancreas is prompted to release insulin and only stops when balance is achieved; hence, negative feedback.

When there is a tear on the skin, a chemical is released. This chemical causes platelets in the blood to activate, hence they release a chemical which signals more platelets to activate, until the wound is clotted, positive feedback.

7 0

3 years ago