Ask question

Ask question

All categories

3 years ago

7

What are 5 reasons to be against Nuclear Power?

1 answer:

emmasim [6.3K]3 years ago

6 0

Well, my 5 reasons would be:

The risk factor. The risk of nuclear power would be INSANE, if we had nukes packed in the backs of our cars and got in a car crash, let's just say we wouldn't have roads after that accident.

Nuclear waste. The waste caused by nuclear is outrageous, it stays for hundreds of thousands of years, and we already dump 2,000 metric tons of nuclear waste per year already, if we had a community based on nuclear, it'd be terrible.

The cost would be tremendous, you would have to build tons of sites, hire probably a new military just for it, scientists, everything would make the cost tremendous.

The foreign risk. Think of all the people who walk into America ( terrorists ) and say "Imma' go blast myself a couple o' Americans". Terrorists would have even more reasons to raid areas for Uranium for the money or to build a bomb, if we got one hit with all the plants we would need, it'd probably cause a very large chain reaction destroying part of our country.

Nuclear radiation. If we have nuclear everything everywhere, how are we gonna protect ourselves from the radiation? We gonna go to work in hazmat suits?

Those are my five reasons, hope it helps, if not, comment below please!!!!

You might be interested in

1. Determine the magnitude of two equal but opposite charges if they attract one another with a force of 0.7N when at distance o

adell [148]

Answer:

q = 2.65 10⁻⁶ C

Explanation:

For this exercise we use Coulomb's law

F = $k \frac{q_1q_2}{r^2}$

In this case they indicate that the load is of equal magnitude

q₁ = q₂ = q

the force is attractive because the signs of the charges are opposite

F = $k \ \frac{q^2}{r^2}$

q = $\sqrt{\frac{F \ r^2}{k} }$

we calculate

q = $\sqrt{\frac{0.7 \ 0.3^2 }{9 \ 10^9} }$

q = $\sqrt{7 \ 10^{-12} }$ Ra 7 10-12

q = 2.65 10⁻⁶ C

7 0

3 years ago

If an object falling freely were somehow equipped with an odometer to measure the distance it travels, then the amount of distan

ioda

Answer:c

Explanation:

Given

object is falling Freely with an odometer

Suppose it falls with zero initial velocity

so distance fallen in time t is given by

$h=ut+\frac{1}{2}gt^2$

here u=0 and t=time taken

$h=\frac{1}{2}gt^2$

for $t=1 s$

$h_1=\frac{1}{2}g$

for $t=2 s$

$h_2=\frac{1}{2}g(2)^2=\frac{4}{2}g=2g$

distance traveled in 2 nd sec $=2g-\frac{1}{2}g=\frac{3}{2}g$

for $t=3 s$

$h_3=\frac{1}{2}g(3)^2=\frac{9}{2}g$

distance traveled in 3 rd sec $=\frac{9}{2}g-2g=\frac{5}{2}g$

so we can see that distance traveled in each successive second is increasing

5 0

3 years ago

Which statement applies only to electric force instead of both electric and magnetic

garri49 [273]

Answer:

C

Explanation:

7 0

3 years ago

I need help on 6, 7, 8, and 9

Ivahew [28]

6 is b. part B on 6 is a. 7 is a. partB ON 7 b

5 0

3 years ago

A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th

NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

$v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2$

$v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)$

$v_{pf}=-31.4\ m/s$

$v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}$

$v_{gf} = 0.0988\ m/s$

4 0

3 years ago