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3 years ago

Which one of these statements true about the moderator and the control rods a nuclear power station?

Physics

1 answer:

siniylev [52]3 years ago

4 0

Answer: the moderator slows the neutrons down, while the control rods absorb the neutrons.

Explanation:

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Why are citizens obligated to respond to such documents?

Zina [86]

To protect the constitutional right to confront ones accused

5 0

3 years ago

A 12-kg piece of metal displaces 1.6 L of water when submerged. Part A Find its density. Express your answer to two significant

Tatiana [17]

Answer:

ρ = 7500 kg/m³

Explanation:

Given that

mass ,m = 12 kg

Displace volume ,V= 1.6 L

We know that

1000 m ³ = 1 L

Therefore V= 0.0016 m ³

When metal piece is fully submerged

We know that

mass = Density x volume

$m=\rho \times V$

Now by putting the values in the above equation

$\rho=\dfrac{12}{0.0016}\ kg/m^3$

ρ = 7500 kg/m³

Therefore the density of the metal piece will be 7500 kg/m³.

6 0

3 years ago

Solve the inequality 2(n+3) – 4<6. Then graph<br> the solution.

Aloiza [94]

The solution is 22 2(n+3)-4&6

6 0

3 years ago

you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and

GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s

46 mph*1.46667=67.4666668 ft/s

$Momentum_B=1125*67.4666668 ft/s$

Initial momentum of car A is given by

$Momentum_A=1515v_a$ where $v_a$ is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

$1515v_a-(1125*67.4666668 ft/s)$

The common velocity is represented as $v_c$ hence after collision, the final momentum is

$Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c$

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

$1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

The acceleration of two cars $a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}$

From kinematic equation

$v^{2}=u^{2}+2as$ hence

$v^{2}-u^{2}=2as$

$0^{2}-(v_c)^{2}=2*-24.1275*19.5$

$v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s$

Substituting the value of $v_c$ in equation $1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

$1515v_a-(1125*67.4666668 ft/s)= 2640*30.67$

$1515v_a=(1125*67.4666668 ft/s)+2640*30.67$

$v_a=\frac {156868.8}{1515}=103.5438 ft/s$

$\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph$

3 0

3 years ago

Which of the following statements are true?

olga nikolaevna [1]

C and d are the right answers

6 0

3 years ago