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babymother [125]
3 years ago
14

Compare the practical uses and limitations of nuclear fission and fusion. Include in your answer a detailed description of the t

wo processes, comparing and contrasting them in terms of the types of atoms involved, the conditions needed to get the reaction started, and the types of products produced.
Physics
1 answer:
shtirl [24]3 years ago
8 0
In fission, an atom is split into two or more smaller, lighter atoms. Fusion, in contrast, occurs when two or more smaller atoms fuse together, creating a larger, heavier atom. For the former, the energy released by fission is a million times greater than that released in chemical reactions, but lower than the energy released by nuclear fusion. On the other hand, fusion process<span> is three to four times greater than the energy released by fission. Hope this would help. Have a nice day.</span>
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A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a
stira [4]

Answer:

Explanation:

Given

acceleration is given by

a=-g-c_Dv^2

where \ddot{y}=a

\dot{y}=v

Also acceleration is given by

a=v\frac{\mathrm{d} v}{\mathrm{d} s}

ds=\frac{v}{a}dv

\int ds=\int \frac{v}{-g-0.001v^2}dv

\Rightarrow Let -g-0.001v^2=t

-0.001\times 2vdv=dt

vdv=-\frac{dt}{0.002}

at\ v_0=50\ m/s,\ t=-g-0.001(50)^2

t=-g-2.5

at v=0,\ t=-g

\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}

\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}

s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}

s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})

s=113.608\ m

when air drag is neglected maximum height reached is

h=\frac{v_0^2}{2g}

h=\frac{50^2}{2\times 9.8}

h=127.55\ m

3 0
3 years ago
Arm ab has a constant angular velocity of 16 rad/s counterclockwise. At the instant when theta = 60
geniusboy [140]

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second.

<h3>How to determine the angular velocity of a collar</h3>

In this question we have a system formed by three elements, the element AB experiments a <em>pure</em> rotation at <em>constant</em> velocity, the element BD has a <em>general plane</em> motion, which is a combination of rotation and traslation, and the ruff experiments a <em>pure</em> translation.

To determine the <em>linear</em> acceleration of the collar (a_{D}), in inches per square second, we need to determine first all <em>linear</em> and <em>angular</em> velocities (v_{D}, \omega_{BD}), in inches per second and radians per second, respectively, and later all <em>linear</em> and <em>angular</em> accelerations (a_{D}, \alpha_{BD}), the latter in radians per square second.

By definitions of <em>relative</em> velocity and <em>relative</em> acceleration we build the following two systems of <em>linear</em> equations:

<h3>Velocities</h3>

v_{D} + \omega_{BD}\cdot r_{BD}\cdot \sin \gamma = -\omega_{AB}\cdot r_{AB}\cdot \sin \theta   (1)

\omega_{BD}\cdot r_{BD}\cdot \cos \gamma = -\omega_{AB}\cdot r_{AB}\cdot \cos \theta   (2)

<h3>Accelerations</h3>

a_{D}+\alpha_{BD}\cdot \sin \gamma = -\omega_{AB}^{2}\cdot r_{AB}\cdot \cos \theta -\alpha_{AB}\cdot r_{AB}\cdot \sin \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \cos \gamma   (3)

-\alpha_{BD}\cdot r_{BD}\cdot \cos \gamma = - \omega_{AB}^{2}\cdot r_{AB}\cdot \sin \theta + \alpha_{AB}\cdot r_{AB}\cdot \cos \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \sin \gamma   (4)

If we know that \theta = 60^{\circ}, \gamma = 19.889^{\circ}, r_{BD} = 10\,in, \omega_{AB} = 16\,\frac{rad}{s}, r_{AB} = 3\,in and \alpha_{AB} = 0\,\frac{rad}{s^{2}}, then the solution of the systems of linear equations are, respectively:

<h3>Velocities</h3>

v_{D}+3.402\cdot \omega_{BD} = -41.569   (1)

9.404\cdot \omega_{BD} = -24   (2)

v_{D} = -32.887\,\frac{in}{s}, \omega_{BD} = -2.552\,\frac{rad}{s}

<h3>Accelerations</h3>

a_{D}+3.402\cdot \alpha_{BD} = -445.242   (3)

-9.404\cdot \alpha_{BD} = -687.264   (4)

a_{D} = -693.867\,\frac{in}{s^{2}}, \alpha_{BD} = 73.082\,\frac{rad}{s^{2}}

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second. \blacksquare

<h3>Remark</h3>

The statement is incomplete and figure is missing, complete form is introduced below:

<em>Arm AB has a constant angular velocity of 16 radians per second counterclockwise. At the instant when θ = 60°, determine the acceleration of collar D.</em>

To learn more on kinematics, we kindly invite to check this verified question: brainly.com/question/27126557

5 0
2 years ago
Andrew is riding his bike is riding in a circle that has a radius of 10 m. He keeps
Dmitrij [34]

The acceleration that Andrew experience  during his ride is 3.6m/s²

The formula for calculating centripetal acceleration is expressed as:

a = v²/r

v is the speed

r is the radius

Given the following expression

v = 6m/s

r = 10m

Substitute the given parameters into the formula

a = 6²/10

a = 36/10

a = 3.6m/s²

Hence the acceleration that Andrew experience  during his ride is 3.6m/s²

Learn more here: brainly.com/question/1268866

8 0
2 years ago
A cube with sides of length 2cm has a mass of 7.36g . calculate the density of the cube
kondor19780726 [428]
Density = 7.36 grams ÷ (2 cm × 2 cm × 2cm) = 0.92 g/cm^3

7 0
3 years ago
Define anterior and posterior in correlation to the body. (ANATOMY)
tatyana61 [14]

Answer:

it should be right it's from go.ogle hm!!!

Explanation:

Anterior or ventral - front (example, the kneecap is located on the anterior side of the leg). Posterior or dorsal - back (example, the shoulder blades are located on the posterior side of the body). Medial - toward the midline of the body (example, the middle toe is located at the medial side of the foot).

8 0
2 years ago
Read 2 more answers
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