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svetoff [14.1K]
3 years ago
14

Suppose 3 neutrons are released by an atom of U-235 in the first stage of a fission chain reaction. Each of these neutrons has e

nough energy to cause another fission reaction in another nucleus of U-235, releasing more neutrons. And if all of those 2nd stage neutrons cause other nuclei to fission in a 3rd stage, how many total U-235 atoms will have undergone fission after these 3 stages of the chain reaction?
A. 27
B. 9
C. 11
D. 12
Physics
2 answers:
8090 [49]3 years ago
5 0
Is not A.) 27 don't know the answer but it is not (A)
DIA [1.3K]3 years ago
4 0
The answer is A.) 27<span />
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What was the speed of a space shuttle that orbited Earth at an altitude of 1482 km?
gregori [183]

Answer:

v = 7121.3 m/s

Explanation:

As we know that the centripetal force for the space shuttle is due to gravitational force of earth due to which it will rotate in circular path with constant speed

so here we will have

\frac{mv^2}{r} = \frac{GMm}{r^2}

here we know that

r = orbital radius = 6370 km + 1482 km

r = 7.852 \times 10^6 m

also we know that

M = 5.97 \times 10^{24} kg

now we will have

v^2 = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{7.852 \times 10^6}

v^2 = 5.07 \times 10^7

v = 7121.3 m/s

3 0
3 years ago
Read 2 more answers
A group of hikers hear an echo 3.3 s after they shout. The temperature is 20◦C.
romanna [79]

Answer:

560 m

Explanation:

The speed of sound in air is approximately:

v ≈ v₀ + 0.6T

where v₀ is the speed of sound at 0°C (273 K) in m/s, and T is the temperature in Celsius.

The speed of sound at 20°C at that altitude is:

v ≈ 327 + 0.6(20)

v ≈ 339 m/s

The sound travels from the hikers to the mountain and back again, so it travels twice the distance.

339 m/s = 2d / 3.3 s

2d = 1118.7 m

d = 559.35 m

Rounding, the mountain is approximately 560 m away.

4 0
3 years ago
A parallel-plate capacitor, made of two circular plates of radius R - 10 cm, is connected in series 2 with a resistor of resista
Lera25 [3.4K]

Answer:

20

Explanation:

8 0
3 years ago
A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi
Mamont248 [21]

Answer:

a=368.97\ m/s^2

Explanation:

Given that,

Initial angular velocity, \omega=0

Acceleration of the wheel, \alpha =7\ rad/s^2

Rotation, \theta=14\ rotation=14\times 2\pi =87.96\ rad

Let t is the time. Using second equation of kinematics can be calculated using time.

\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s

Let \omega_f is the final angular velocity and a is the radial component of acceleration.

\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s

Radial component of acceleration,

a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2

So, the required acceleration on the edge of the wheel is 368.97\ m/s^2.

6 0
3 years ago
A star has an absolute magnitude of 4 and a surface temperature of 5,000 degrees C. According to the HR diagram, list the type o
Ierofanga [76]
On sources it says it would  just be the super giant star 
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3 years ago
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