All categories

4 years ago

The greatest problem facing the use of nuclear power plants is:

Physics

2 answers:

Dominik [7]4 years ago

7 0

How to dispose of and safely store the active fission products in the spent fuel

levacccp [35]4 years ago

6 0

Answer:

1) Accidents in the nuclear plants

2) Disposal of waste materials released from this power plants.

Explanation:

The use of nuclear power plant is quite efficient, but it contains radioactive elements that are highly radioactive and there occurs continuous emission of radiations that are harmful for health. There is no release of any green house gas and fossil fuels are not burnt.

The problems associated with the use of nuclear power plant are as follow-

Accidents in the plants- In these nuclear power plants, operations are carried out in a very careful manner, because the use of these radioactive elements releases harmful radiations, and accidents happens unexpectedly.
Disposal of waste materials- This nuclear plants releases a huge amount of waste materials that contains some amount of radioactive toxic element that are exposed to the environment causing harm to the various life forms. So it is still not solved how to handle and store this waste products.

You might be interested in

constant force F=5i+5j−1kF=5i+5j−1k is applied to an object that is moving along a straight line from the point (−5,−3,−4)(−5,−3

meriva

Answer:

W = 71J

Explanation:

Given force F = (5i+5j−1k)N

d = Δr

r1 = (−5,−3,−4)m

r2 = (2,5,0)m

Δr = r2 – r1 = (2-(-5), 5-(-3), 0-(-4))

Δr = (2+5, 5+3, 0+4) = (7i+ 8j +4k)m

W = F•d = (5i+5j−1k)•(7i+ 8j +4k)

W = 5×7 + 5×8 +-1×4 = 35 + 40 - 4

W = 71J

6 0

3 years ago

A series circuit has a capacitor of 10−5 F, a resistor of 3 × 102 Ω, and an inductor of 0.2 H. The initial charge on the capacit

Alexus [3.1K]

Given the values to proceed to solve the exercise, we resort to the solution of the exercise through differential equations.

The problem can be modeled through a linear equation, in the form:

$10^5 Q +300Q'+0.2Q''=0$

With the initial conditions as,

$Q(0) = 10^{-6}$

$Q'(0)= 0$

Where Q(t) is the charge.

The general solution of a linear equation is given as:

$y(x) = c_1e^{-ax}+c_2e^{-bx}$

Applying this definiton in our differential equation we have that

$Q(t) = C_1e^{at}+C_2e^{bt}$

To find b and a we use the first equation and find the roots:

$r_{a,b} = \frac{-300 \pm \sqrt{(300)^3-4(0.2)*10^5}}{0.4}$

$r_{a,b} = {-1000,-500}$

Then we have

$Q(t) = C_1e^{-1000t}+C_2e^{-500t}$

To find the values of the Constant we apply the initial conditions, then

$Q(0)= 10^{-6} = C_1+C_2$

And for the derivate:

$Q'(t) = -1000C_1e^{-1000t}-500C_2e^{-500t}$

$0 = -1000C_1e^{-1000(0)}-500C_2e^{-500(0)}$

$0 = -1000C_1-500C_2$

We have a system of 2x2:

$(1) 10^{-6} = C_1+C_2$

$(2) 0 = -1000C_1-500C_2$

Solving we have:

$C_1 = -10^{-6}$

$C_2 = 2*10^{-6}$

The we can replace at the equation and we have that the Charge at any moment is given by,

$Q(t) = (-10^{-6})e^{-1000t}+( 2*10^{-6})e^{-500t}$

If we obtain the derivate we find also the Current, then

$I(t)= 10^{-3}e^{-1000t}-10^{-3}e^{-500t}$

7 0

3 years ago

What are the name of variables that do not change through out and experiment?

kifflom [539]

Answer:

The answer is C

Explanation:

3 0

3 years ago

Read 2 more answers

Car A of mass 1200kg traveling at 10m/s , collided in too the back of the car B, which is stationary. Following the collision,.

Nina [5.8K]

Answer:

Car B has a mass of 800 kg.

General Formulas and Concepts:

Momentum

Law of Conservation of Momentum: $\displaystyle m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_{f}$

Explanation:

Step 1: Define

Identify variables

[Given] m₁ = 1200 kg

[Given] v₁i = 10 m/s

[Solve] m₂

[Given] v₂i = 0 m/s

[Given] vf = 6 m/s

Step 2: Solve for m₂

Substitute in variables [Law of Conservation of Momentum]: (1200 kg)(10 m/s) + m₂(0 m/s) = (1200 kg + m₂)(6 m/s)
Multiply: 12000 kg · m/s = (1200 kg + m₂)(6 m/s)
Isolate m₂ term: 2000 kg = 1200 kg + m₂
Isolate m₂: 800 kg = m₂

4 0

3 years ago

Read 2 more answers

An isotope has a half life of 2.5 minutes how many half -lives go by in 17.5 minutes

ale4655 [162]

Answer:

7 half-lives

Explanation:

7 0

4 years ago