Answer and Explanation:
The aluminum is more productive in the absorption and heat transfer to other particles. It instantly converts heat absorbed from the environment into the atmosphere when removed from the oven, enabling us to operate with it faster than the pie that takes much longer to convert heat to the environment.
So this is the reason for pie to be the dangerously hot
The rate of fuel burning in grams per hour if the DT reaction is used is 1.08 ×
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×
×1.6×
) ÷ ( 5.030×1.66×
)
= 3.37×
J/Kg
= 3.0× 
J/g
Rate of fuel burning in grams per hour = 3.0× × 3600
= 3.6×3.0×
= 1.08 × J/g per hour
To learn more about fusion reactor and energy production, refer
brainly.com/question/13399644
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AAaaahh yes, T'was that a lovely chap by the name of Copernicus, has discovered us a little secret, well. why dont you share?
Copernicus: well uh, i'm a bit shy
ME: who a gives a flying *bleep* about your feelings man! just man up an discover the earth's rotation already!
Answer:
i got b but urs is a little different tell me if right
Explanation:
i go 100 on my test
<span>11.823 cm
There is a slight ambiguity with this question in that I don't know if the measurements are from the surface of the ball, or the center of the ball. I will take this question literally and as such the point light source will be 124 cm from the wall.
The key thing to remember is that ball won't be showing an effective diameter of 4 cm to the light source. Instead the shadow line is a tangent to the ball's surface. There is a right triangle where the hypotenuse is the distance from the center of the ball to the light source (42 cm), one leg of the triangle is the radius (2cm). That right triangle will define a chord that will be the effective diameter of the disk casting the shadow. The cosine of the half angle of the chord will be 2/42 = 1/21. The sine of the half angle then becomes sqrt(1-(1/21)^2) = sqrt(440/441) = 2sqrt(110) = 0.99886557. Now multiply that sine by 4 (radius of ball multiplied by 2 since it's the half angle and we want the full side of the chord) and we get an effective diameter of 3.995462279 cm.
Now we need to calculate the effective distance that circle is from the wall. It will be slightly larger than 82 cm. The exact value will be 82 + cos(half angle) * radius. So
82 + 1/21 * 2 = 82 + 2/21 = 82.0952381
Now we have the following dimensions with a circle replacing the ball in the original problem.
Distance from wall to effective circle = 82.0952381 cm
Distance from effective circle to point source = 124 - 82.0952381 = 41.9047619 cm
Effective diameter of circle = 3.995462279 cm
And because the geometry makes similar triangles, the following ratio applies.
3.995462279/41.9047619 = X/124
Now solve for X
3.995462279/41.9047619 = X/124
124*3.995462279/41.9047619 = X
495.4373226/41.9047619 = X
11.82293611 = X
The shadow cast on the wall will be a circle with a diameter of 11.823 cm</span>
