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1 year ago

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(a) Consider a fusion generator built to create 3.00 GW of power. Determine the rate of fuel burning in grams per hour if the DT

reaction is used.

1 answer:

Stels [109]1 year ago

4 0

The rate of fuel burning in grams per hour if the DT reaction is used is 1.08 × $10^1^3$ J/g per hour

<h3>How is the rate of fuel burning in grams per hour calculated when the D-T reaction is used?</h3>

D + T → He + n
The D-T fusion reaction results in a Helium (He) and neutron (n)

E = 17.59 MeV

Mass = 2.014u + 3.016u

= 5.030u

Energy per Kg = (17.59× $10^6$ ×1.6× $10^-^1^9$ ) ÷ ( 5.030×1.66× $10^-^2^7$ )

= 3.37× $10^1^4$ J/Kg

= 3.0× $10^9$ J/g

Rate of fuel burning in grams per hour = 3.0× $10^9$ × 3600

= 3.6×3.0× $10^1^2$

= 1.08 × $10^1^3$ J/g per hour

To learn more about fusion reactor and energy production, refer

brainly.com/question/13399644

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2 years ago

A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

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Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

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$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

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here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

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$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

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Part b)

Now we know that speed of the ball after collision is given as

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so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

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2 years ago

A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m

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Answer:

The magnitude of the vector $\vec{C}$ is 107.76 m

Explanation:

To find the components of the vectors we can use:

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where $| \vec{A} |$ is the magnitude of the vector, and θ is the angle over the positive x axis.

The negative x axis is displaced 180 ° over the positive x axis, so, we can take:

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$\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )$

$\vec{B} = ( \ -53.797 \ m \ , \ 61.886\ m \ )$

Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

$(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)$

So, for our vectors:

$\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ , ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )$

$\vec{C} = ( \ - 102.294 \ m , \ 33.886 m \ )$

To find the magnitude of this vector, we can use the Pythagorean Theorem

$|\vec{C}| = \sqrt{C_x^2 + C_y^2}$

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And this is the magnitude we are looking for.

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2 years ago

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Answer:

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2 years ago

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