Answers to electrical?

What is a disadvantage of using nuclear power to produce electricity?

MrRa [10]

Question:

What is a disadvantage of using nuclear power to produce electricity?

Answer:

Disadvantages of Nuclear Power

The further implementations of nuclear power are limited because although nuclear energy does not produce CO2 the way fossil fuels do, there is still a toxic byproduct produced from uranium-fueled nuclear cycles: radioactive fission waste.

7 0

3 years ago

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Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. Wha

ikadub [295]

We have that the spring constant is mathematically given as

$k=2.37*10^{11}N/m$

Generally, the equation for angular velocity is mathematically given by

$\omega=\sqrt{k}{m}$

Where

k=spring constant

And

$\omega =\frac{2\pi}{T}$

Therefore

$\frac{2\pi}{T}=\sqrt{k}{n}$

Hence giving spring constant k

$k=m((\frac{2 \pi}{T})^2$

Generally

Mass of earth $m=5.97*10^{24}$

Period for on complete resolution of Earth around the Sun

$T=365 days$

$T=365*24*3600$

Therefore

$k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2$

$k=2.37*10^{11}N/m$

In conclusion

The effective spring constant of this simple harmonic motion is

$k=2.37*10^{11}N/m$

For more information on this visit

brainly.com/question/14159361

8 0

3 years ago

Water flows through a multisectional pipe placed horizontally on the ground. The velocity is 3.0 m/s at the entrance and 2.1 m/s

Stolb23 [73]

Answer:

$P_1 - P_2 = 2295 pascal$

Explanation:

As we know by Bernoulli's principle

$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$

$P_1 - P_2 = \frac{1}{2}\rho v_2^2 - \frac{1}{2} \rho v_1^2$

here we know that

$\rho = 1000 kg/m^3$

also we know that

$v_1 = 2.1 m/s$

$v_2 = 3 m/s$

now we have

$P_1 - P_2 = \frac{1}{2}(1000)(3.0^2 - 2.1^2)$

$P_1 - P_2 = 2295 pascal$

5 0

3 years ago

A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top

Anton [14]

Answer:

$y_{max} = 0.829\,m$

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

$v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}$

$v = 2.913\,\frac{m}{s}$

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

$(3\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m) + \frac{1}{2}\cdot (3\,kg)\cdot (2.913\,\frac{m}{s} )^{2} = (3\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s) ^{2}$

After some algebraic handling, a second-order polynomial is formed:

$12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} - (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s$

$1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0$

The roots of the polynomial are, respectively:

$\Delta s_{1} \approx 0.128\,m$

$\Delta s_{2} \approx -0.099\,m$

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

$\Delta s \approx 0.128\,m$

The maximum height that the block reaches after rebound is:

$(3\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s)^{2} = (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot y_{max}$

$y_{max} = 0.829\,m$

4 0

3 years ago

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A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates with a constant upward acceleration

tamaranim1 [39]

In the question where as a model rocket is launched straight upward with an initial speed of 50.0m/s and it accelerate with a constant acceleration of 2m/s^2 until it engines stop at an altitude od 150. So the following are the answer to your question:
a.The motion, the acceleration was equals to Zero
b.the maximum height is 150m

8 0

3 years ago

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