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3 years ago

What shape is the JET experimental fusion reactor?

Physics

2 answers:

Sauron [17]3 years ago

7 0

Answer:

doughnut-shaped chamber called the tokamak. This is where the fusion reactions take place, within hot plasma containing deuterium and tritium atoms.

Karo-lina-s [1.5K]3 years ago

3 0

The shape of JET experimental fusion reactor is the 'D-shaped' vacuum chamber.

<u>Explanation: </u>

JET, “Joint European Torus” is World's biggest magnetically compact plasma physics experiment, in Oxfordshire, UK. The tokamak design paved way to the future nuclear fusion grid energy.

JET was the first tokamaks designed to use this D-shaped chamber. The initial construction of this chamber was purely based on safety factor, but later it was noted that the system was built mechanically to reduce the net force across the chamber which forces the torus towards middle of major axis.

To support the forces, magnets around the chamber should be curved at top and bottom and less on inside to support the acting forces leads to oval shaped chamber. The D shape of the reactor has the dimension of about 2.5 meters in width and 4.2 meters in height with the total of 100 cubic meters of volume for the plasma.

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Butoxors [25]

Answer:

Explanation:

Given

mass of bus along with travelers travelling in North direction is $m_1=1.6\times 10^4 kg$

speed of bus towards North $v_1=15.2 km/h\approx 4.22\ m/s$

mass of bus travelling in South direction is $m_2=1.578\times 10^4 kg$

speed of bus $v_2=12.2 km/h\approx 3.38\ m/s$

mass of each Passenger in south moving bus $m_0=64.8 kg$

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$P_1=m_1\times v_1$

$P_1=1.6\times 10^4\times 4.22$

$P_1=6.768\times 10^4 kg-m/s$

Momentum with south moving bus

$P_2=m_2\times v_2+n\cdot m_0\times v_2$

$P_2=(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38$

For total momentum to be towards south

$P_2-P_1$ should be greater than 0

thus for least value of n

$P_2=P_1$

$(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38=6.768\times 10^4$

$1.578\times 10^4+n\cdot 64.8=2.0023\times 10^4$

$n=\frac{4243.6686}{64.8}=65.48\approx 66$

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3 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

lyudmila [28]

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$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
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By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
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part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
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marusya05 [52]

Answer:

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Explanation:

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An archer shoots an arrow 75 m distant target; the bull's-eye of the target is at the same height as the release height of the a

Liono4ka [1.6K]

Answer:

$16.25^{\circ}$

Explanation:

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Horizontal range is given by

$R=\dfrac{v^2\sin2\theta}{g}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{Rg}{v^2}}{2}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{75\times 9.81}{37^2}}{2}\\\Rightarrow \theta=16.25^{\circ}$

The angle at which the arrow is to be released is $16.25^{\circ}$ .

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