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anastassius [24]
2 years ago
6

What total mass must be converted into energy

Physics
2 answers:
Eduardwww [97]2 years ago
7 0

This question apparently wants you to get comfortable
with  E = m c² .  But I must say, this question is a lame
way to do it.

c = 3 x 10⁸ m/s
                                                    E = m c²

                           1.03 x 10⁻¹³ joule  =  (m) (3 x 10⁸ m/s)²

Divide each side by (3 x 10⁸ m/s)²:

                         Mass = (1.03 x 10⁻¹³ joule) / (9 x 10¹⁶ m²/s²)

                                   =  (1.03 / 9) x (10⁻¹³ ⁻ ¹⁶) (kg)

                                   =        1.144 x 10⁻³⁰  kg .    (choice-1)

This is roughly the mass of (1 and 1/4) electrons, so it seems
that it could never happen in nature.  The question is just an
exercise in arithmetic, and not a particularly interesting one.
______________________________________

Something like this could have been much more impressive:
 
The Braidwood Nuclear Power Generating Station in northeastern
Ilinois USA serves Chicago and northern Illinois with electricity.
<span>The station has two pressurized water reactors, which can generate
a net total of 2,242 megawatts at full capacity, making it the largest
nuclear plant in the state.
If the Braidwood plant were able to completely convert mass
to energy, how much mass would it need to convert in order
to provide the total electrical energy that it generates in a year,
operating at full capacity ?

Energy = (2,242 x 10⁶ joule/sec) x (86,400 sec/day) x (365 da/yr)

             =  (2,242 x 10⁶ x 86,400 x 365) joules

             =          7.0704 x 10¹⁶ joules .

How much converted mass is that ?

                                           E  =  m c²

Divide each side by  c² :    Mass  =  E / c² .
c = 3 x 10⁸ m/s

              Mass = (7.0704 x 10¹⁶ joules) / (9 x 10¹⁶ m²/s²)

                        =        0.786 kilogram ! ! !

THAT should impress us !  If I've done the arithmetic correctly,
then roughly  (1 pound  11.7 ounces) of mass, if completely
converted to energy, would provide all the energy generated
by the largest nuclear power plant in Illinois, operating at max
capacity for a year !

</span>
Lorico [155]2 years ago
5 0
     The equivalence between mass and energy is given by:

E=mc^2
  
     Entering the unknowns, we have:

1.03*10^{-13}=m*(3*10^8)^2 \\ m= \frac{1.03*10^{-13}}{9*10^{16}}  \\ \boxed {m=1.14*10^{-30}Kg}

Number 1

If you notice any mistake in my english, please let me know, because i am not native.
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Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of south.
Nina [5.8K]

Answer:

The resultant vector \vec R = \vec A+\vec B is given by \vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m].

Explanation:

Let \vec A = 6\cdot (\cos 30^{\circ}\,\hat{i}+\sin 30^{\circ}\,\hat{j}) and \vec B = 4\cdot (-\sin 30^{\circ}\,\hat{i}-\cos 30^{\circ}\,\hat{j}), both measured in meters. The resultant vector \vec R is calculated by sum of components. That is:

\vec R = \vec A+\vec B (Eq. 1)

\vec R = 6\cdot (\cos 30^{\circ}\,\hat{i}+\sin 30^{\circ}\,\hat{j})+4\cdot (-\sin 30^{\circ}\,\hat{i}-\cos 30^{\circ}\,\hat{j})

\vec R = (6\cdot \cos 30^{\circ}-4\cdot \sin 30^{\circ})\,\hat{i}+(6\cdot \sin 30^{\circ}-4\cdot \cos 30^{\circ})\,\hat{j}

\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]

The resultant vector \vec R = \vec A+\vec B is given by \vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m].

8 0
3 years ago
A hoop of mass 2 kg, radius 0.5 m is rotating about its center with an angular speed of 3 rad's. A force of 10N is applied tange
Degger [83]

Answer:

The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.

Explanation:

Given that,

Mass = 2 kg

Radius = 0.5 m

Angular speed = 3 rad/s

Force = 10 N

(I). We need to calculate the rotational kinetic energy

Using formula of kinetic energy

K.E =\dfrac{1}{2}\timesI\omega^2

K.E=\dfrac{1}{2}\times mr^2\times\omega^2

K.E=\dfrac{1}{2}\times2\times(0.5)^2\times(3)^2

K.E=2.25\ J

(II). We need to calculate the instantaneous change rate of the kinetic energy

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

On differentiating

\dfrac{K.E}{dt}=\dfrac{1}{2}m\times2v\times\dfrac{dv}{dt}

\dfrac{K.T}{dt}=mva....(I)

Using newton's second law

F = ma

a= \dfrac{F}{m}

a=\dfrac{10}{2}

a=5 m/s^2

Put the value of a in equation (I)

\dfrac{K.E}{dt}=mva

\dfrac{K.E}{dt}=mr\omega a

\dfrac{K.E}{dt}=2\times0.5\times3\times5

\dfrac{K.E}{dt}=15\ J/s

Hence, The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.

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