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3 years ago

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Multiple-Concept Example 6 reveiws the principles that play a role in this problem. A nuclear power reactor generates 2.3 x 109

W of power. In one year (365.25 days), what is the change in the mass of the nuclear fuel due to the energy being taken from the reactor?

2 answers:

Alika [10]3 years ago

6 0

Answer:

The change in the mass of the nuclear fuel due to the energy being taken from the reactor is 0.81 kg

Explanation:

Given:

P = power 2.3x10⁹W

The energy taking from the reactor is:

E = P * t = 2.3x10⁹ * 365 * 24 * 60 * 60 = 7.25x10¹⁶J

The change in the mass is:

E = Δm * c²

Where c is speed of light in vacuum

Δm = E/c² = 7.25x10¹⁶/(3x10⁸)² = 0.81 kg

r-ruslan [8.4K]3 years ago

3 0

Answer:

change in mass = 2.41*10^{8}kg

Explanation:

The change in the mass can be computed by using the relation

$E=\Delta mc^2\\\Delta m=\frac{E}{c^2}$ (1)

That is, the energy liberated comes from the mass of the nuclear fuel. The energy generated in one year is

$E=Pt=2.3*10^{9}\frac{J}{s}*1 year*\frac{365.25 day}{1 year}*\frac{24h}{1 day}*\frac{3600s}{1h}=7.25*10^{16}J$

Hence, by replacing in the equation (1) you have (c=3*10^{8}m/s)

$\Delta m=\frac{7.25*10^{16}J}{3*10^{8}\frac{m}{s}}=2.41*10^{8}kg$

HOPE THIS HELPS!!

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Answer:

The ratio of their orbital speeds are 5:4.

Explanation:

Given that,

Mass of A = 5 m

Mass of B = 7 m

Radius of A = 4 r

Radius of B = 7 r

The orbital speed of satellite A,

$v_{A}=\sqrt{\dfrac{GM_{A}}{R_{A}}}$ ......(I)

The orbital speed of satellite B,

$v_{B}=\sqrt{\dfrac{GM_{B}}{R_{B}}}$ ......(I)

We need to calculate the ratio of their orbital speeds

Using equation (I) and (II)

$\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{\dfrac{GM_{A}}{R_{A}}}{\dfrac{GM_{B}}{R_{B}}}}$

Put the value into the formula

$\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{G\times5m\times7r}{G\times7m\times4r}}$

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Hence, The ratio of their orbital speeds are 5:4.

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cricket20 [7]

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The kinetic energy of an object with a mass of 6.8 kg and a velocity of 5.0 m/s is J. (Report the answer to two significant figu

abruzzese [7]

Answer:

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Given:

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To Find:

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Explanation:

Formula:

$\boxed{ \bold{\sf KE = \frac{1}{2} m {v}^{2} }}$

Substituting values of m & v in the equation:

$\sf \implies KE = \frac{1}{2} \times 6.8 \times {5}^{2}$

$\sf \implies KE = \frac{1}{ \cancel{2}} \times \cancel{2} \times 3.4 \times 25$

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3 years ago

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The magnitude of the force associated with the gravitational field is constant and has a value FF . A particle is launched from

uysha [10]

Answer:

The kinetic energy of the particle will be 12U₀

Explanation:

Given that,

A particle is launched from point B with an initial velocity and reaches point A having gained U₀ joules of kinetic energy.

Constant force = 12F

According to question,

The kinetic energy is

$U_{0}=Fx$ ....(I)

Constant force = 12F

A resistive force field is now set up ,

Resistive force is given by,

$F_{r}=12F$

When the particle moves from point B to point A then,

We need to calculate the kinetic energy

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$U=F_{r}x$

Put the value of $F_{r}$

$U=12Fx$

Now, from equation (I)

$U=12U_{0}$

Hence, The kinetic energy of the particle will be 12U₀.

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3 years ago

The intensity at distance from a spherically symmetric sound source is 100 W/m2. What is the intensity at five times this distan

ss7ja [257]

To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.

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So replacing we have to

$I = \frac{P}{4\pi r^2}$

Since the question tells us to find the proportion when

$r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}$

So considering the two intensities we have to

$I_1 = \frac{P_1}{4\pi r_1^2}$

$I_2 = \frac{P_2}{4\pi r_2^2}$

The ratio between the two intensities would be

$\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}$

The power does not change therefore it remains constant, which allows summarizing the expression to

$\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2$

Re-arrange to find $I_2$

$I_2 = I_1 (\frac{r_1}{r_2})^2$

$I_2 = 100*(\frac{1}{5})^2$

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