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sladkih [1.3K]
3 years ago
11

During a heat wave, there is an increased demand for electricity to operate air conditioners. The basic design of a nuclear reac

tor that generates electricity is shown below. What is the best way for the power plant manager to modify the reactor to meet the demand for electricity?
a)lower the control rods, which will absorb more neutrons before they can split atoms
b)raise the control rods so that fewer neutrons are absorbed and more can split uranium atoms
c)remove some of the fuel rods, which will leave more room for additional neutrons to split atoms
d)add additional fuel rods to generate additional neutrons to split uranium atoms
Physics
2 answers:
Lelechka [254]3 years ago
4 0
Given that the design is not shown, for traditional nuclear reactors with control rods, the best way to meet an increased demand for energy would be to raise the control rods so that less neutrons are absorbed.

So, the most reasonable answer to the question is B.
nlexa [21]3 years ago
3 0

Answer:

b)raise the control rods so that fewer neutrons are absorbed and more can split uranium atoms

Explanation:

Here in nuclear reactor we know that high energy neutrons are projected on uranium atom which will further break the uranium into smaller atoms and thus energy is released.

In this whole process the reaction and the released energy is controlled by the help of control rods which will absorb the high speed neutrons and control the energy of the reaction.

Now if energy demand is increased then in this case the energy demand can be fulfilled by generating more energy for which we can remove the control rods so that more number of neutrons will strike with uranium atom and hence more energy will release.

so correct answer will be

b)raise the control rods so that fewer neutrons are absorbed and more can split uranium atoms

You might be interested in
A nonconducting rod of length L = 8.15 cm has charge –q = -4.23 fC uniformly distributed along its length.(a) What is the linear
vichka [17]

Answer:

a)  λ = 5.19 10⁻⁴ C/m , b)  E = 1,573 10⁻³ N/C , c) the direction of the field is directed to the bar

Explanation:

a) the linear density defined as the ratio between the charge per unit length

       λ = q / l

Let's start by reducing the units to the SI system

     L = 8.15 cm (1m / 100cm) = 8.15 10⁻² m

     a = 12 cm (1m / 100cm) = 12 10⁻² m

    q = -4.23 fC (1 C / 10¹⁵ ft) = -4.23 10⁻¹⁵ C

    λ = -4.23 10⁻¹⁵ C / 8.15 10⁻²

    λ = 5.19 10⁻⁴ C/m

b) Let's look for the electric field for a point at a distance a from the end of the bar

      E = k  dq / r²

To simplify the notation, suppose the bar is the x axis. Since the density is constant, we can write it differentially

     λ = dq/dx

     dq = λ dx

     E = k ∫ λ dx / x²

We integrate and evaluate between the lower limits x = a and higher x = L + a. Here we place the test point at the origin of the system

     E = k λ (-1 / x)

     E = k λ (-1 /(L + a) + 1 /a)

     E = k λ (L /a(L + a)

Let's change the density for its value

     E = k (q / L) (L / a (L + a)

     E = k q  1 /[a(L + a)]

     E = 8.99 10⁹ 4.23 10⁻¹⁵ [1 /12 10⁻²(8.15 10⁻² + ​​12 10⁻²)]

     E = 1,573 10⁻³ N/C  

c) the direction of the field is directed to the bar, because it has a negative charge

d) now we change the distance a = 50 cm = 0.50 m

Bar

      E = 8.99 10⁹ 4.23 10⁻¹⁵ ( 1 /0.5(0.0815 +0.5))

      E = 1,308 10⁻⁴ N/C

Charge point

      q = -4.23 10⁻¹⁵ C

     E = k q / r²

     E = 8.99 10⁹ 4.23 10⁻¹⁵ / 0.5²

     E = 1.521 10⁻⁴ N/C

7 0
3 years ago
What is the correct water depth for an echo travel time of 6 seconds? (for the purposes of this exercise, assume pressure and te
saul85 [17]
In physics, there are empirical values for common important parameters. For example, the speed of sound is equal to 340 meters per second. Unlike the speed of light, the speed of sound is dependent on temperature and pressure. But in standard room conditions, the speed is 340 m/s. Using this value of speed, we can find the depth given the time. You should also note that the distance the echo travels is exactly the same distance that object travelled. Therefore,

Distance = 340 m/s * 6 seconds
Distance = 2,040 meters deep
7 0
4 years ago
Which of the followings did newton help create ?
vladimir1956 [14]
I think the anwser is 3)calculus
4 0
3 years ago
Read 2 more answers
A parachutist bails out and freely falls 63 m. Then the parachute opens, and thereafter she decelerates at 1.5 m/s2. She reaches
yaroslaw [1]

Answer:

(a) The parachutist spent 24.84 secs in air

(b) The height the fall begins is 472 m

Explanation:

Here is the complete question:

A parachutist bails out and freely falls 63 m. Then the parachute opens, and thereafter she decelerates at 1.5 m/s2. She reaches the ground with a speed of 3.3 m/s. (a) How long is the parachutist in the air  (b) At what height does the fall begin?

Explanation:

From one of the equations of kinematics for free fall

H = ut - \frac{1}{2}gt^{2}

Where H is the height

u is the initial velocity

t is the time

and g is the acceleration due to gravity (Take g = 9.8 m/s2)

Now, we can find the time spent before the parachute opens.

u = 0 m/s (we assume the parachutist starts from rest)

H = - 63 m

∴-63 = 0(t) - \frac{1}{2}(0.98) t^{2}  \\-63 = -4.9 t\\t^{2} = \frac{63}{4.9} \\t^{2} = 12.86\\t = \sqrt{12.86} \\t = 3.59 s

This the time spent before the parachute opens

Also from one of the equations of kinematics for free fall

v = u - gt

where v is the final velocity

We can determine the final velocity before the parachute opens and she starts to decelerate

∴v = - 9.8(3.59)\\v = - 35.18m/s

Now, we will calculate the time spent after the parachute opens

From one of the equation of kinematics for linear motion,

v = u + at

Here, the initial velocity will be the final velocity just before the parachute opens, that is

u = - 35.18 m/s

From the question,

v = - 3.3 m/s

a = 1.5 m/s^{2}

We then get

-3.3 = - 35.18 + (1.5)t

-3.3 + 35.18 =  1.5t\\31.88 = 1.5t

t = \frac{31.88}{1.5}

t = 21.25 secs

(a) To determine how long the parachutist is in the air,

That is sum of the time used when falling freely and the time used after the parachute opens

= 3.59 secs + 21.25 secs

24.84 secs

Hence, the parachutist spent 24.84 secs in air

(b) To determine what height the fall begins

First, we will calculate the height from which the parachute opens

From one of the equation of kinematics for linear motion,

x = ut + \frac{1}{2}at^{2}  \\

x = -35.18(21.25) + \frac{1}{2}(1.5)(21.25)^{2}  \\x = -747.58 + 338.67\\x = -408.91m\\

x ≅ - 409m

Hence, the height the fall begins is 63m + 409m

= 472 m

3 0
3 years ago
A guitar string is 0.620m long, and oscillates at 234Hz. What is the velocity of the waves in the string? m/s
9966 [12]

Answer:

v = 72.54 m/s

Explanation:

We have,

Length of a guitar string is 0.62 m

Frequency of a guitar string is 234 Hz

For guitar string,

L=2\lambda\\\\\lambda=\dfrac{L}{2}\\\\\lambda=\dfrac{0.62}{2}\\\\\lambda=0.31\ m

The velocity of the wave in the string is given by :

v=f\lambda\\\\v=234\times 0.31\\\\v=72.54\ m/s

So, the velocity of the waves in the string is 72.54 m/s.

3 0
4 years ago
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