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2 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]2 years ago

5 0

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

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What is the efficiency of a machine?

Julli [10]

Answer:

Efficiency is the percent of work put into a machine by the user (input work) that becomes work done by the machine (output work).

Explanation:

It is a measure of how well a machine reduces friction.

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2 years ago

A 4.0-m-diameter playground merry-go-round, with a moment of inertia of

HACTEHA [7]

Answer:

$7.1$ ms⁻¹

Explanation:

$d$ = diameter of merry-go-round = 4 m

$r$ = radius of merry-go-round = $\frac{d}{2}$ = $\frac{4}{2}$ = 2 m

$I$ = moment of inertia = 500 kgm²

$w_{i}$ = angular velocity of merry-go-round before ryan jumps = 2.0 rad/s

$w_{f}$ = angular velocity of merry-go-round after ryan jumps = 0 rad/s

$v$ = velocity of ryan before jumping onto the merry-go-round

$m$ = mass of ryan = 70 kg

Using conservation of angular momentum

$Iw_{i} - m v r = (I + mr^{2})w_{f}$

$(500)(2.0) - (70) v (2) = (I + mr^{2})(0)$

$1000 = 140 v$

$v = 7.1$ ms⁻¹

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2 years ago

Explain how buoyancy can defeat the force of gravity

Olenka [21]

Answer:

Buoyancy causes things to float in water or liquid which appears to defeat the force of gravity. Hope this helps.

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2 years ago

A hockey ball accelerates from 0.m/s to 25m/s in 0.05 seconds what is the acceleration of the ball ?

S_A_V [24]

Answer:

500

Explanation:

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I think this ?

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2 years ago

Air is cooled in a process with constant pressure of 150 kPa. Before the process begins, air has a specific volume of 0.062 m^3/

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Answer:

The pressure is constant, and it is P = 150kpa.

the specific volumes are:

initial = 0.062 m^3/kg

final = 0.027 m^3/kg.

Then, the specific work can be written as:

$W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.$

The fact that the work is negative, means that we need to apply work to the air in order to compress it.

Now, to write it in more common units we have that:

1 kPa*m^3 = 1000J.

-5.25 kPa*m^3/kg = -5250 J/kg.

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3 years ago