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Mama L [17]
3 years ago
10

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?
Physics
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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Dmitry [639]

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}}

\dfrac{v_{A}}{v_{B}}=\dfrac{5}{4}

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

8 0
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cricket20 [7]
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5 0
2 years ago
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:

\boxed{\sf Kinetic \ energy \ (KE) = 85 \ J}

Given:

Mass (m) = 6.8 kg

Speed (v) = 5.0 m/s

To Find:

Kinetic energy (KE)

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

\sf \implies KE =3.4 \times 25

\sf \implies KE = 85 \: J

8 0
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

Using formula for kinetic energy

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.

7 0
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.

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

I = \frac{P}{A}

The area of a sphere is given by

A = 4\pi r^2

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

I_2 = 4W/m^2

Therefore the intensity at five times this distance from the source is 4W/m^2

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3 years ago
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