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2 years ago

Why are nuclear power plants controversial

1 answer:

8 0

I would have to say because while they do produce relatively clean power, it is extremely dangerous and overly harmful should one "melt down." Some people believe the reward is worth the risk, others do not. It is more a matter of perspective - are you near the nuclear power plant or not? If you are, chances are you won't like it - on the other hand, if you are not right hear the power plant you might like it more. I hope this helps!

P.S. - If you think I did a good job I would love it if you went to my YouTube channel to see if I can help you more there. I am just starting out and don't have much time to record, but I would love to have some user-submitted topics! Here is the link: https://www.youtube.com/channel/UCG0zQcSsxZ2mu-OnZRdI2-Q. I hope to see you there!

You might be interested in

What is the frequency of the electromagnetic wave if the period is 1.54x10-15s

pogonyaev

answer✿࿐

I was not able to write it here

so I did it somewhere else and attached the picture

i hope it helps

have a nice day

#Captainpower

3 0

2 years ago

A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p

777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

$v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s$

Now, we can calculate the angular acceleration (w0=0rad/s)

$\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}$

$\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}$

with this value we can compute the angular velocity

$\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}$

and the tangential velocity of point B, and then the acceleration of point B:

$v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}$

hope this helps!!

6 0

2 years ago

Read 2 more answers

Which electromagnetic wave has wavelengths just longer than visible light?

vlabodo [156]

Infra-red has wavelengths just longer than visible light.

7 0

2 years ago

Read 2 more answers

He formula p=m x v to answer th<br>2,000 kg car moving at 30 m/s

aivan3 [116]

The momentum of the car is 60,000 kg m/s

Explanation:

The momentum of an object is a vector quantity, whose magnitude is given by

$p=mv$

where

m is the mass of the object

v is its velocity

For the car in this problem, we have

m = 2,000 kg

v = 30 m/s

Therefore, the magnitude of its momentum is

$p=(2000)(30)=60,000 kg m/s$

Learn more about momentum here:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

7 0

2 years ago

To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where

BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

$\int E.dA = \frac{q}{\epcilon_0}$

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

$q = \int \rho dV$

$q = \rho * \pi r^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r}{2 \epcilon_0}$

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

$q = \int \rho dV$

$q = \rho * \pi r_0^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r_0^2}{2 \epcilon_0 r}$

7 0

2 years ago