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Usimov [2.4K]
3 years ago
5

Nuclear fusion has not yet been harnessed as an energy source because

Physics
2 answers:
Katen [24]3 years ago
7 0

Answer:

B

Explanation:

Energy is hard to control.

maksim [4K]3 years ago
6 0

Before getting an answer for it first we have to understand nuclear fusion.

Nuclear fusion is a thermo-nuclear reaction in which two  light unstable nuclei will form a heavy stable nuclei. In this process there will be some mass defect which will be converted into energy as per Einstein's mass energy equivalence theorem.

The theorem is stated as E = mc^{2} where c is the velocity of light and m is the mass converted into energy.

One take an example of fusion in sun where 4 hydrogen atoms combine to form a helium nucleus which are explained below-

                                                      H_{1}^{1}+H_{1} ^{1} =H_{1} ^{2} +e_{+1} ^{0} +energy

                                                      H_{1} ^{2} +H_{1} ^{1} =He_{2} ^{3} +energy

                                                      He_{2} ^{3}+H_{1} ^{1} =He_{2} ^{4} +e_{+1} ^{0} +energy

                                                       -----------------------------------------------------------------------

                                                        4H_{1} ^{1} = He_{2} ^{4} +2e_{+1} ^{0} +energy

Here e_{+1} ^{0} is the positron.

In this process very high temperature is needed which is approximately equal to the temperature of the sun i.e 10^{6} K

Such temperature is very difficult to initiate the reaction on the earth surface. It should be carried out in an sustainable way also .Otherwise It will cause nuclear hazards.

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A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g
sweet-ann [11.9K]

Answer:

(a) h = 1.27 Re

(b) h = 1.86 Re

Explanation:

Let M is the mass of earth and Re is the radius of earth.

initial velocity of projection, v = 0.462 ve

where, ve is the escape velocity of an object on earth surface.

(a)

The value of escape velocity is

v_{e}=\sqrt{\frac{2GM}{R_{e}}}

So, v=0.462\times \sqrt{\frac{2GM}{R_{e}}}    .... (1)

By using conservation of energy

(Kinetic energy + potential energy ) at the surface of earth = Potential energy at the height h.

where, h is the maximum height upto which the projectile reach

K.E at surface  P.E at surface = P.E at the top

\frac{1}{2}mv^{2}-\frac{GMm}{R_{e}}=-\frac{GMm}{h}

By equation (1), substituting the value of v

\frac{1}{2}\times 0.462^{2}\times \frac{2GM}{R_{e}}-\frac{GM}{R_{e}}=-\frac{GM}{h}

\frac{1}{2}\times 0.462^{2}\times \frac{2}{R_{e}}-\frac{1}{R_{e}}=-\frac{1}{h}

h = 1.27 Re

(b)

initial kinetic energy = 0.462 times the kinetic energy required to escape

\frac{1}{2}mv^{2}=0.462\times \frac{1}{2}mv_{e}^{2}

\frac{1}{2}mv^{2}=0.462\times \frac{1}{2}m\times \frac{2GM}{R_{e}}

So, again by using the conservation of energy

Kinetic energy at the surface + Potential energy at the surface = Potential energy at the top

0.462\times \frac{1}{2}m\times \frac{2GM}{R_{e}}-\frac{GMm}{R_{e}}=-\frac{GMm}{h}

0.462\times \frac{1}{R_{e}}-\frac{1}{R_{e}}=-\frac{1}{h}

h = 1.86 Re

7 0
3 years ago
_SeC16+_O2>_SeO2+_C12
juin [17]

Answer:

SeC14

Explanation:

sec14>C12-O

but since you can't subtract o from c , you get this

7 0
3 years ago
In a women's 100-m race, accelerating uniformly, Laura takes 1.82 s and Healan 3.07 s to attain
Sav [38]

Answer:

Laura is ahead and for a distance of 3.22 m

Explanation:

To solve this problem of one-dimensional kinematics, we have to find the acceleration and the final speed of each runner. Let's start with Laura

Lura data is acceleration time 1.82s, total run time 10.4 s and total distance 100m.  In all the races the  rest starts, so the initial speed is zero (Vo = 0)

   Vf1= Vo + a1 t1    

   Vf1 = x/t                

   XT  = X1 + X2

   X1 = Vo t1 + ½ a1 t1²  

   X1 = ½ a1 t1²  

   X2 = Vf1 (t-t1)

This is the remaining time of the race after the acceleration is over.

    XT = ½ a1 t1² + Vf1 (t-t1)

We remplace the expression of Vf1

     XT = ½ a1 t1² + a1 t1 (t-t1)

Laura's aceleration (a1) is

   a1= XT / [ ½  t1² + t1 (t-t1)]

   a1= 100/ [ ½ 1.82²+ 1.82 (10.4 -1.82)]

   a1=  5.79m/s2  

We repeat the same calculation for the other Healan runner, whose data are: total distance 100m, acceleration time 3.07 s and total time 10.4 s

Vf2= Vo + a2 t2    

Vf2 = x/t                

XT  = X3 + X4

X3 = Vo t2 + ½ a2 t2²  

X3 = ½ a2 t2²  

X4 = Vf2 (t-t2)

XT = ½ a2 t2² + Vf2 (t- t2)

XT = ½ a2 t2² + a2 t2 (t-t2)

The aceleration of Healan (a2)

a2 = XT / [½ t2² + t2 (t-t2)]

a2 = 100 / [½ 3,07²+ 3.07 (10.4 -3.07)]

a2 = 3.67 m / s2

We also need the final speeds of each runner

Laura Vf1 = Vo + a1 t1

          Vf1 = 0 + 5.79 1.82

          Vf1 = 10.54 m / s

Healan Vf2 = Vo + a2 t2

            Vf2 = 0 + 3.67 3.07

            Vf2 = 11.27 m / s

Having the acceleration and speed of each runner, you can start answering the questions

a) For t3 = 6.15s

Laura

The time to stop with constant speed is what remains after accelerating

XL= ½ a1 t1² + Vf1 (t3-t1)

XL= ½ 5.79 1.82² + 10.54 (6.15 – 1.82)    

XL= 55.23 m

Healan  

XH= ½ a2 t2² + Vf2 (t3-t2)

             XH= ½  3.67 3.07² + 11.27 (6.15-3.07)

             XH= 52.01 m

             (XL -XH)= 55.23- 52.01

             (XH -XL)=  3.22 m

It is appreciated from these results that Laura is ahead and for a distance of 3.22 m

b) If we analyze the acceleration values ​​of each runner, knowing that they leave the rest and that Healan at the end has a speed greater than Laura, the point of maximum distance difference is when Laura stops accelerating t = 1.82 s

      XL= ½  a1 t12  

      XL= ½ 5.79 1.822

      XL= 9.59 m

      XH = ½ a2 t12

      XH= ½ 3.67 1.822

      XH= 6.08 m

The maximum distance difference is 3.51 m

c) Already analyzed in the previous part 1.82 s, since the Laura stop accelerating and Heala continue with acceleration will travel greater distances in equal time units

6 0
3 years ago
What is the mass of one sheet of printer paper?
Sloan [31]
A standard sheet of laser-printer paper of the usual thickness, either US "letter" size or international "A4" size, weighs 4.5 grams. That means 100 sheets weighs almost exactly one pound.
4 0
3 years ago
A teacher performing demonstration finds that a piece of cork displaces 23.5 ml of water. The piece of cork has a mass 5.7 g. Wh
Sedbober [7]

Answer: 0.24g/ml

Explanation:

Given that:

Volume of water displaced = 23.5 ml

Mass of cork = 5.7 g

Density of the cork = ?

Recall that density is obtained by dividing the mass of a substance by the volume of water displaced.

i.e Density = Mass/volume

Density = 5.7g /23.5ml

Density = 0.24g/ml

Thus, the density of the piece of cork is 0.24g/ml

8 0
4 years ago
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