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hammer [34]
3 years ago
12

A reactor operating at 1 MW is scrammed by instantaneous insertion of $5.00 of negative reactivity. Approximately how long does

it take the power to drop to 1 milliwatt?
Physics
1 answer:
Novosadov [1.4K]3 years ago
4 0

Answer:

time is 3333.33 min or 55.55 hr

Explanation:

given data

reactor operating = 1 MW

negative reactivity =  $5

power = 1 miliwatt

to find out

how long does it take

solution

we know here power coefficient that is

power coefficient = \frac{10^{6} }{10^{6} }

power coefficient = 1

so time required to reach power is

power =  reactivity × time / power coefficient + reactor operating

1 × 10^{-3} = -5 t / 1  +  1 × 10^{6}

5t =  10^{6} - 10^{-3}

t = 199999.99 sec

so time is 3333.33 min or 55.55 hr

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A ball is launched from ground level at 20 m/s at an angle of 40° above the
DedPeter [7]

(a) The ball's height <em>y</em> at time <em>t</em> is given by

<em>y</em> = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve <em>y</em> = 0 for <em>t</em> :

0 = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²

0 = <em>t</em> ((20 m/s) sin(40º) - 1/2 <em>g t</em> )

<em>t</em> = 0   or   (20 m/s) sin(40º) - 1/2 <em>g t</em> = 0

The first time refers to where the ball is initially launched, so we omit that solution.

(20 m/s) sin(40º) = 1/2 <em>g t</em>

<em>t</em> = (40 m/s) sin(40º) / <em>g</em>

<em>t</em> ≈ 2.6 s

(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration <em>g</em>. So

0² - ((20 m/s) sin(40º))² = 2 (-<em>g</em>) <em>y</em>

where <em>y</em> in this equation refers to the maximum height of the ball. Solve for <em>y</em> :

<em>y</em> = ((20 m/s) sin(40º))² / (2<em>g</em>)

<em>y</em> ≈ 8.4 m

8 0
3 years ago
A 2. 0 μf and a 4. 0 μf capacitor are connected in series across an 8. 0-v dc source. what is the charge on the 2. 0 μf capacito
Nezavi [6.7K]

voltage across 2.0μf capacitor is 5.32v

Given:

C1=2.0μf

C2=4.0μf

since two capacitors are in series there equivalent capacitance will be

[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]

c =  \frac{c1 \times c2}{c1 + c2}

=  \frac{2 \times 4}{2 + 4}

=1.33μf

As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.

Q=CV

given,V=8v

= 1.33 \times 10 {}^{ - 6}  \times 8

= 10.64 \times 10 {}^{ - 6}

charge on 2.0μf capacitor is

\frac{Qeq}{2 \times 10 {}^{ - 6} }

=  \frac{10.64 \times 10 {}^{ - 6} }{2 \times 10 {}^{ - 6} }

=5.32v

learn more about series capacitance from here: brainly.com/question/28166078

#SPJ4

3 0
2 years ago
In Millikan's oil drop experiment an oil drop is at rest between two large plates separated by 1.3 cm when the potential differe
sweet-ann [11.9K]

Answer:

Mass of the oil drop, m=3.01\times 10^{-15}\ kg

Explanation:

Potential difference between the plates, V = 400 V

Separation between plates, d = 1.3 cm = 0.013 m

If the charge carried by the oil drop is that of six electrons, we need to find the mass of the oil drop. It can be calculated by equation electric force and the gravitational force as :

qE=mg

m=\dfrac{qE}{g}

q=6e, e is the charge on electron

E is the electric field, E=\dfrac{V}{d}=\dfrac{400}{0.013}=30769.23\ V/m

m=\dfrac{6\times 1.6\times 10^{-19}\times 30769.23}{9.8}

m=3.01\times 10^{-15}\ kg

So, the mass of the oil drop is 3.01\times 10^{-15}\ kg. Hence, this is the required solution.

5 0
3 years ago
Scientists have changed the model of the atom as they have gathered new evidence. One of the atomic models is shown below. Overl
olga55 [171]

Answer:

A). A few of the positive particles aimed at a gold foil seemed to bounce back off of the thin metallic foil.

Explanation:

Scientists decided to change the model of the atom when they discovered new evidence that showed 'few of the positive particles aimed at a gold foil seemed to bounce back off of the thin metallic foil.' On this ground, <u>Rutherford concluded that atom is mostly made up of empty space and thus, he proposed a nucleus model of atom in which the atom comprises of the tiny and positively charged nucleus is surrounded by electrons with a negative charge</u>. Thus, <u>option A</u> is the correct answer.

8 0
3 years ago
Read 2 more answers
Which statement is true about the technology used to improve the motion of space vehicles on rough surfaces?
Andrei [34K]

The correct answer is C.)

It has made road vehicles safer because magnetometers are used to detect particles found in radiation emitted during combustion of fuel.

h a v e  a  g r e a t  d a y

4 0
3 years ago
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