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Neporo4naja [7]
3 years ago
8

Four heavy elements (A, B, C, and D) will fission when bombarded by neutrons. In addition to fissioning into two smaller element

s, A also gives off a beta particle, B gives off gamma rays, C gives off neutrons, and D gives off alpha particles. Which element would make a possible fuel for a nuclear reactor
Physics
1 answer:
Zolol [24]3 years ago
5 0

Answer:

<em>Element C will be best for a nuclear fission reaction</em>

Explanation:

<em>Nuclear fission is the splitting of the nucleus of a heavy atom by bombarding it with a nuclear particle. The reaction leads to the the atom splitting into two smaller elements and a huge amount of energy is liberated in the process.</em> For the reaction to be continuous in a chain reaction,<em> the best choice of element to use as fuel for the reaction should be the element whose nucleus also liberates a neutron particle after fission</em>. The neutron that is given off by other atoms in the reaction will then proceed to bombard other atoms of the element in the reaction, creating a cascade of fission and bombardment within the nuclear reactor.

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The process of beta emission can be envisioned as the.
Ivan

Answer:

conversion or a neutron to a proton and electron. the electronic is emitted

6 0
3 years ago
A car traveling at 27 m/s runs out of gas while traveling up a 8.0 ∘ slope. How far will it coast before starting to roll back d
Ad libitum [116K]

Answer:

Approximately 2.7 \times 10^{2}\; \rm m along the slope, assuming that no energy was lost to friction.

Explanation:

Let v denote the initial velocity of this vehicle. Let m denote the mass of this vehicle. The kinetic energy (KE) of this vehicle would have initially been:

\displaystyle \frac{1}{2}\, m \cdot v^{2}.

The gain in the gravitational potential energy (GPE) of this vehicle is proportional to the increase in its height.

Let g denote the gravitational field strength. g \approx 9.81\; \rm m \cdot s^{2} on the earth. If the increase in the height of this vehicle is h, this vehicle would have gained GPE:

m \cdot g \cdot h.

Hence, the height of this vehicle is maximal when the GPE of this vehicle is maximized.

Since the vehicle went out of fuel, all its GPE would have been converted from KE. Assuming that no energy was converted to friction. The GPE of this vehicle would be maximal when the entirety of the KE was converted to GPE.

Hence, the maximal GPE of this vehicle would be equal to its initial KE:

\displaystyle m \cdot g \cdot h = \frac{1}{2}\, m \cdot v^{2}.

The maximum height of this vehicle would be:

\begin{aligned} h &= \frac{(1/2) \, m \cdot v^{2}}{m \cdot g}\\ &= \frac{v^{2}}{2\, g}\end{aligned}.

Given that v = 27\; \rm m\cdot s^{-1}, the maximum height of this vehicle would be:

\begin{aligned} & \frac{v^{2}}{2\, g} \\ =\; & \frac{(27\; \rm m\cdot s^{-1})^{2}}{2 \times 9.81\; \rm m\cdot s^{-2}} \\ \approx \; &37.2\; \rm m\cdot s^{-1}\end{aligned}.

Refer to the diagram attached. The distance that this vehicle traveled along the slope would be approximately:

\begin{aligned}\frac{37.2\; \rm m}{\sin(8.0^{\circ})} \approx 2.7 \times 10^{2}\; \rm m\end{aligned}.

6 0
3 years ago
Consider an infinetly long hollow cylindrical wire. The wire has an outer diameter b and the clindrical hole at its center has d
Nadusha1986 [10]

Answer:

B = (μ₀*i/(2*π*x))*(x²-(a/2)²)/((b/2)²-(a/2)²)

Explanation:

Given

Outer diameter of the wire = b  ⇒ R = b/2

Diameter of the clindrical hole at the center = a  ⇒ r = a/2

The current that flows from left to right and is uniformly spread over the region between a and b = i

We apply Ampere's Law

Using the following formula for  a/2 ≤ x ≤ b/2

B = (μ₀*i/(2*π*x))*(x²-(a/2)²)/((b/2)²-(a/2)²)

6 0
3 years ago
Projectiles move in which two directions?
AfilCa [17]

Analyzing two-dimensional projectile motion is done by breaking it into two motions: along the horizontal and vertical axes.

hope it helps!!

5 0
3 years ago
Read 2 more answers
A 0.4-kg cart, traveling on a horizontal air track with a speed of 6 m/s, collides with a stationary 0.8-kg cart. The carts stic
steposvetlana [31]

Answer:

B.1.6 N*s

Explanation:

According to the principle of conservation of momentum, we have:

\Delta p=0\\p_f-p_i=0\\p_f=p_i\\v_fm_f=m_1v_1+m_2v_2

The final mass is obtained adding the masses of the two cars since they stick together after the collision. So, m_f=m_1+m_2. Recall that the 0.4 kg cart collides with the stationary 0.8-kg cart. So v_2=0

v_f(m_1+m_2)=m_1v_1\\v_f=\frac{m_1v_1}{m_1+m_2}\\v_f=\frac{0.4kg(6\frac{m}{s})}{0.4kg+0.8kg}\\v_f=2\frac{m}{s}

The magnitude of the impulse is defined as the mass multiplied by the change in the speed:

I=m\Delta v \\I=m(v_f-v_i)\\I=0.8kg(2\frac{m}{s}-0)\\I=1.6N\cdot s

3 0
4 years ago
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