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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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2 years ago
A system gains 1500 J of heat, while the internal energy of the system increases by 4500 J and the volume decreases by . Assume
Assoli18 [71]

Answer:

Hence the pressure is 3\times 10^5 Pa

Explanation:

Given data

Q=1500 J   system gains heat

ΔV=- 0.010 m^3     there is a decrease in volume

ΔU= 4500 J        internal energy decrease

We know work done is

W= Q- ΔU

=1500-4500= -3000 J

The change in the volume at constant pressure is

ΔV= W/P

there fore P = W/ΔV= -3000/-0.01= 3×10^5

Hence the pressure is 3\times 10^5 Pa

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2 years ago
How to find average speed
marshall27 [118]

Answer:

Average speed (s) = total distance/total elapsed tiime

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8 0
3 years ago
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A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.
gtnhenbr [62]

Answer:

(a)  vmax = 0.34m/s

(b)  v = 0.13m/s

(c)  v = 0.31m/s

(d)  x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

v_{max}=\omega A       (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

\omega=\sqrt{\frac{k}{m}}           (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

x=Acos(\omega t)

You solve the previous equation for t:

t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s

With this value of t, you can calculate the speed of the object with the following formula:

v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s

The position will be:

x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m

The position of the object on which its speed is one-half its maximum velocity is 0.039

5 0
3 years ago
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