Ask question

Ask question

All categories

3 years ago

11

During test-out procedures, a 239Pu-fueled thermal reactor is operated for a time at a power of 1 megawatt.The power is then to

be increased to 100 megawatts in 8 hours. (a) On what stable period should the reactor be placed? (b) What reactivity insertion is required?

1 answer:

miskamm [114]3 years ago

5 0

The stable placed on x+7*8 the equation for your hysics will be solved high iron paid

You might be interested in

A 5 kilogram block is dropped from a height of 3 meters and falls straight to the ground. What is the work done by the force of

vazorg [7]

Answer:

Workdone = 147Nm

Explanation:

Given the following data;

Mass = 5kg

Height = 3m

We know that acceleration due to gravity is 9.8m/s²

First of all, we would find the force applied;

$Force = mass * acceleration$

Force = 5*9.8

Force = 49N

Now, to find the workdone;

$Workdone = force * distance$

Substituting into the equation, we have;

Workdone = 49 * 3

Workdone = 147Nm

5 0

3 years ago

Byron bought a keyless entry door lock that has the digits 0 through 9 on the keypad. He wants to choose a three-digit entry cod

DedPeter [7]

Answer:

27

Explanation:

8 0

4 years ago

Read 2 more answers

Two violinists are trying to tune their instruments in an orchestra. One is producing the desired frequency of 440.0 hz. The oth

Katyanochek1 [597]

Answer:

Percentage change in tension is 3.8%

Explanation:

We have given initially frequency $f_1$ = 440 Hz

Let tension in the string at this frequency is $T_1$

Now second frequency is $f_2=448.4Hz$

Frequency in string is given by

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$

From the relation we can say that

$\frac{f_1}{f_2}=\sqrt{\frac{T_1}{T_2}}$

$\frac{440}{448.4}=\sqrt{\frac{T_1}{T_2}}$

${\frac{T_2}{T_1}}=1.038$

Percentage change in tension is equal to

$=\frac{T_2-T_1}{T_1}=\frac{T_2}{T_1}-1=(1.038-1)\times 100=3.8$ %

So percentage change in tension is 3.8%

4 0

3 years ago

What is Potentiometer

Katena32 [7]

Answer:

an instrument for measuring an electromotive force by balancing it against the potential difference produced by passing a known current through a known variable resistance.

3 0

3 years ago

A vertical wire carries a current straight up in a region where the magnetic field vector points due north. What is the directio

Elanso [62]

Answer:

The direction of the resulting force on this current is due east.

Explanation:

Given;

direction of the magnetic field to be due north

Applying right hand rule which states that: to determine the direction of the magnetic force on a positive moving charge point the thumb of the right hand in the direction of velocity v, the fingers in the direction of magnetic field B, and a perpendicular to the palm points in the direction of magnetic force.

Since the magnetic force must be perpendicular to the magnetic field, and direction of the magnetic field is due north, then the magnetic force must be due East.

Therefore, the direction of the resulting force on this current is due east.

7 0

3 years ago