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3 years ago

Describe three risks or disadvantages of nuclear power.

Physics

2 answers:

Galina-37 [17]3 years ago

4 0

Answer:

due to the properties of radioactive elements like penetration power and ionization effect they can cause gene mutations, kill living cells and cause cancer. They also have the advantages. They are used to make nuclear energy

jok3333 [9.3K]3 years ago

3 0

Answer:

Fuel for nuclear power plants is a radiation hazard to miners and other nuclear power workers. Nuclear waste requires long-term storage. Mining nuclear fuels can damage the environment. Human error at plants can lead to meltdowns, which endanger the environment and human health. Natural disasters can lead to radiation leaks. Though currently abundant, nuclear fuels are nonrenewable resources.

Explanation:

this is the sample answer :)

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How do very small objects behave?

nataly862011 [7]

A. very small objects behave like like particles.

7 0

4 years ago

Read 2 more answers

Light shined through a single slit will produce a diffraction pattern. Green light (565 nm) is shined on a slit with width 0.210

kondor19780726 [428]

Answer:(a)9.685 mm

(b)4.184 mm

Explanation:

Given

Wavelength of light $(\lambda )=565nm \approx 565\times 10^{-9}m$

Width of slit(b)=0.210

(a)Width of central maximum located 1.80m from slit

$=\frac{2\lambda L}{b}$

$=\frac{2\times 565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}$

=9.685 mm

(b)Width of the first order bright fringe

$Y_1=\frac{\lambda \times L}{b}$

$Y_1=\frac{565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}$

$Y_1=4.84mm$

5 0

3 years ago

An 87.6 g lead ball is dropped from rest from a height of 7.00 m. The collision between the ball and the ground is totally inela

bija089 [108]

Taking specific heat of lead as 0.128 J/gK = c

We have energy of ball at 7.00 meter height = mgh = $87.6*10^{-3}*9.81*7$

When leads gets heated by a temperature ΔT energy needed = mcΔT

= $87.6*10^{-3}*0.128*10^3$ ΔT

Comparing both the equations

$87.6*10^{-3}*9.81*7$ = $87.6*10^{-3}*0.128*10^3$ ΔT

ΔT = 0.536 K

Change in temperature same in degree and kelvin scale

So ΔT = 0.536 $^0C$

7 0

3 years ago

What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

Wittaler [7]

Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$