To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volume V the ratio of the sphere will be
.
We know that the surface area and volume of the sphere is given by:

Therefore, the ratio between the surface area and the volume for the sphere will be:

Equating the volume to the constant c, we will find the value of
.

Substituting the value of r in the ration between surface area and volume, we get:

Calculating the constants, we get:

Hence, the ration between surface area and volume is 
To learn more about surface area and volume of sphere, refer to:
brainly.com/question/4387241
#SPJ4
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
vA (0) = 0^2 – 5(0) + 20
vA = 20 m/s
For vB
vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
Learn more: brainly.com/question/24857760
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?
Answer: The color is orange, the state of matter is liquid
Explanation:
Answer:
0.75 m
Explanation:
Let's call the distance between the bulb and the mirror x.
The bulb and the length of the mirror form a triangle. The mirror and the illuminated area on the floor form a trapezoid. If we extend the lines from the mirror edge to the reflected image of the bulb, we turn that trapezoid into a large triangle. This triangle and the small triangle are similar. So we can say:
x / 0.4 = (3 + x) / 2
Solving for x:
2x = 0.4 (3 + x)
2x = 1.2 + 0.4 x
1.6 x = 1.2
x = 0.75
So the bulb should located no more than 0.75 m from the mirror.
Answer:
The frequency of sound wave created by trumpet is 437.5Hz
Explanation:
Given
the speed of sound wave = 350 m
the wavelength of sound wave = 0.8 m
the frequency of sound wave = ?
All the waves have same relationship among wavelength, frequency and speed, which is given by the equation:
v = fλ, where
v is speed of the wave
f is frequency of the wave
λ is wavelength of the wave
therefore frequency of sound wave is given by
f = v/λ
= 350m
/0.8m
= 437.5
= 437.5Hz (since 1
= 1 Hz (Hertz)
Hence the frequency of sound wave created by trumpet is 437.5Hz