Since fusion and fission are opposite processes that both produce energy,why can we not simply run the process forward and then backwardrepeatedly and have a limitless supply of energy?A. The products of a fission reaction cannot be used for a fusionreaction, and the products of a fusion reaction cannot be used fora fission reaction.B. Fusion reactions can occur cheaply enough, but fission requiresvery high temperatures.C. Fusion produces energy from nuclei larger than iron, and fissionproduces energy from nuclei smaller than iron.D. Fission reactions can occur cheaply enough, but fusion requires very high temperatures

Answers

Answer 1

Answer:

ANSWER:

D. Fission reactions can occur cheaply enough, but fusion requires very high temperatures

STEP-BY-STEP EXPLANATION:

One of the main reasons fusion power cannot be harnessed is that its power requirements are incredibly high. For fusion to occur, a temperature of at least 100,000,000°C is needed.

Therefore, the correct answer is D. Fission reactions can occur cheaply enough, but fusion requires very high temperatures

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A conducting sphere has a net charge of -4.8x10-17 C. What is the approximate number of excess electrons on the sphere?
A spring has a spring constant of 1350 N/m. You place the spring vertically with one end on the floor. You then drop a 1.3 kg book onto it from a height of 0.8 m above the top of the spring. Find the maximum distance the spring will be compressed. Express your answer with the appropriate mks units.

If a ball leaves the ground with a vertical velocity of 5.46 m/s, how long does it takethe ball to reach the highest point?

Answers

Answer:

0.557 s

Explanation:

Given:

v₀ = 5.46 m/s

v = 0 m/s

a = -9.8 m/s²

Find: t

v = at + v₀

0 m/s = (-9.8 m/s²) t + 5.46 m/s

t = 0.557 s

A tightly wound solenoid is 15 cm long, has 350 turns, and carries a current of 4.0 A. If you ignore end effects, you will find that the value of app at the center of the solenoid when there is no core is approximately

Answers

Answer:

The magnetic field at the center of the solenoid is approximately 0.0117 T

Explanation:

Given;

length of the solenoid, L = 15 cm = 0.15 m

number of turns of the solenoid, N = 350 turns

current in the solenoid, I = 4.0 A

The magnetic field at the center of the solenoid is given by;

$B = \mu_o ((N)/(L) )I\n\nB = (4 \pi *10^(-7))((350)/(0.15) )(4.0)\n\nB = 0.0117 \ T$

Therefore, the magnetic field at the center of the solenoid is approximately 0.0117 T.

The interatomic spring stiffness for tungsten is determined from Young's modulus measurements to be 90 N/m. The mass of one mole of tungsten is 0.185 kg. If we model a block of tungsten as a collection of atomic "oscillators" (masses on springs), what is one quantum of energy for one of these atomic oscillators? Note that since each oscillator is attached to two "springs", and each "spring" is half the length of the interatomic bond, the effective interatomic spring stiffness for one of these oscillators is 4 times the calculated value given above. Use these precise values for the constants: ℏ = 1.0546 10-34 J · s (Planck's constant divided by 2π) Avogadro's number = 6.0221 1023 molecules/mole kB = 1.3807 10-23 J/K (the Boltzmann constant)

Answers

Answer:

Explanation:

solution below

Final answer:

The quantum of energy for one atomic oscillator in tungsten, given the effective interatomic spring stiffness of 360 N/m, the mass of one tungsten atom as 3.074 x 10^-25 kg, and the reduced Planck's constant of 1.0546 x 10^-34 J · s, can be calculated to be approximately 1.33 x 10^-21 J.

Explanation:

To calculate the quantum of energy for one atomic oscillator in tungsten, we will consider the model of an atom being connected to two springs, both having an effective interatomic spring stiffness of four times the given value (90 N/m). This value thus becomes 360 N/m.

One mole of tungsten has a mass of 0.185 kg, thus the mass of one atom can be determined by dividing this value by Avogadro's number (6.0221 x 10^23 molecules/mole), which gives approximately 3.074 x 10^-25 kg.

The quantum of energy, or the energy of one quantum (the smallest possible energy increment), is given by the formula E = ħω, where ħ is the reduced Planck's constant (1.0546 x 10^-34 J · s) and ω is the angular frequency, given by sqrt(k/m), where k is the spring constant and m is the mass.

Substituting the known values into these equations gives ω= sqrt((360)/(3.074 x 10^-25)) and E= (1.0546 x 10^-34) x sqrt((360)/(3.074 x 10^-25)), which results in a quantum of energy of approximately 1.33 x 10^-21 J.

Learn more about Quantum Energy here:

brainly.com/question/28175160

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The ____ contains the highest concentration of ozone

Answers

I believe the term you are looking for is the ozone layer. This layer is in the highest region if the stratosphere.

A 50n brick is suspended by a light string from a 30kg pulley which may be considered a solid disk with radius 2.0m. the brick is released from rest and falls to the floor below as the pulley rotates. it takes 4 seconds for the brick to hit the floor. i) what is the tension in newtons in the string well the brick is falling? ii) what is the magnitude of the angular momentum in kg*m^2/s of the pulley at the instant the brick hits the floor?

Answers

Brick is held at a position which is at height 2 m from the floor

Now it is released from rest and hit the floor after t = 4 s

Now the acceleration of the brick is given by

$y = v_i* t + 0.5 at^2$

$2 = 0 + 0.5 * a * 4^2$

$a = 0.25 m/s^2$

Now in order to find the tension in the string

we can use Newton's law

$F_(net) = ma$

$mg - T = ma$

$50 - T = (50)/(9.8)*0.25$

$T = 48.72 N$

part b)

Now for the pulley

moment of inertia= $(1)/(2)mr^2$

m = 30 kg

R = 2 m

I = $(1)/(2)*30*2^2$

I = 60 kg m^2

Now the angular speed just before brick collide with the floor

$w = (v)/(r)[\tex]</p><p>here we have</p><p>[tex]v = v_i + a* t$

$v = 0 + 0.25 * 4$

v = 1 m/s

Now we will have

L = angular momentum = I w = $I*(v)/(R)$

L = 60 * $(1)/(2)$

L = 30 kg m^2/s

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

Answers

Answer:

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_(1) +q_(2) =-q_(3)$ -----eqution 1

where,

$q_(1)$ is the heat absorbed by the solid at 0⁰C

$q_(2)$ is the heat absorbed by the liquid at 0⁰C

$q_(3)$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

$=237g *(1 mole H_(2) O)/(18g) = 13.167 moles of H_(2)O$

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_(1) = 13.167 moles *6.01(KJ)/(mole) = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_(2) = -q_(3)$

Step 4: calculate the final temperature of the water

$79.13KJ+M_(sample) *C*\delta {T_(sample)} =-M_(water) *C*\delta {T_(water)$

Substitute in the values; we will have,

$79.13KJ + 237*4.18(J)/(g^(o)C)*(T_(f)-218}) = -350*4.18(J)/(g^(o)C)*(T_(f)-100})$

79.13 kJ + 990.66J* $(T_(f)-218})$ = -1463J* $(T_(f)-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_(f)-218})$ = -1.463KJ* $(T_(f)-100})$

$79.13 + 0.99066T_(f) -215.96388= -1.463T_(f)+146.3$

collect like terms,

2.45366 $T_(f)$ = 283.133

∴ $T_(f) =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

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