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Ghella [55]
3 years ago
6

Which best describes fusion nuclear reactors? They use rods containing uranium for fuel. They are currently used for electricity

in several highly populated countries. They are used in some naval vessels, such as submarines. They have been successful, but they are very expensive to operate.
Physics
2 answers:
Ludmilka [50]3 years ago
6 0
They has been very successful but they are very expensive to operate that is your answer I hope this helps
andrew-mc [135]3 years ago
5 0
<h2>Right answer: They have been successful, but they are very expensive to operate. </h2>

A fusion reactor is a facility in which nuclear fusion reactions take place in a fuel formed by hydrogen isotopes (deuterium and tritium), releasing energy in the form of heat, and then transforming it into electrical energy.

At the moment there is no commercial fusion reactor that allows to obtain electrical energy, although there are experimental research facilities where fusion reactions are studied, as well as the technology that will be used in the future.

In other words:

All commercial nuclear reactors currently in use work with <u>nucler fission</u> (separation of the components of the atom nucleous), not <u>nuclear fusion</u> (joining two light nucleous to form a heavier nucleous). And this is because, although experimental reactors have already shown us that nuclear fusion works, it is such a complex process that has not been yet found a way to get the fusion reaction to behave in a stable way for a long time and it is expensive as well.


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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:

A)

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

So

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

B)

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

C)

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

D)

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

E)

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

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3 years ago
While on a playground, you and your niece take turns sliding down a frictionless slide. Your mass is 75 kg while your little nie
Olenka [21]

Answer:

c.

Explanation:

  • In absence of friction, total mechanical energy must be conserved.
  • This means that the change in gravitational potential energy (in magnitude) must be equal to the change in kinetic energy:

       \Delta K = \Delta U \\ \\  \frac{1}{2} * m* v^{2}  = m*g*h

  • As it can be seen, the mass m is on the both sides of the equation, which means that it can be simplified.
  • We can solve this equation for v (speed at the bottom) as follows:

        v = \sqrt{2*g*h}

  • As it can be seen, the mass has no part in the equation, so, due both are starting from the same height, both people have the same speed at the bottom.
  • The option c is the right one.
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3 years ago
Potential energy is the ability to:
Ymorist [56]

Answer:

potential energy is the ability to do work

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Which layer lies between the stratosphere and the
Pepsi [2]
The answer is Mesosphere
8 0
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A colloidal liquid in a has is called
S_A_V [24]
Aerosol is the answer
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