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Andrews [41]
3 years ago
6

What are 3 pros and cons of nuclear transmutation?

Physics
2 answers:
Kay [80]3 years ago
8 0
Pros It doesn't cost much Power stations are ver compact Transportation of the material is easy Cons It isn't safe Nuclear waste take 200 years to degrade Greater risk of explosion.
Tanya [424]3 years ago
6 0
Pros -

It doesn't cost much
Power stations are very compact
Transportation of the material is easy

Cons-

It's not safe
Nuclear waste take 200 years to degrade
Greater risk of explosion



      -IF THIS HELPED WHICH I HOPE COMMENT BELOWw ↓PLEASEX . THANKs!
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How many grams of CO-60 result in 1 Millicuire of activity? How many years until the activity decays to 1 microcuire tl/2 =5.3 Y
vredina [299]

Answer:

m =8.81*10^{-6}grams

time t = 52.8 year

Explanation:

GIVEN DATA:

the half life of the CO-60 is, T_1/2 = 5.27 years = 1.663 e+8 s

activity dN/dt = 1 mCi = 3.7 X 10^7 decay/s

activity ,  dN/dt = N* \lambda

              \frac{dN}{dt} = N* \frac{ln2}{T_1/2}

                   N = \frac{(dN/dt )(T1/2)}{ln2}

                      = ( 3.7 X 10^7 )(1.663*10^8 ) / ln2

                      = 8.877*10^{16}

Number of moles:

     n = N/NA = 8.877*10^{16} / 6.022X10^23 = 1.474*10^{-7} mol

mass of the CO-60 is,

   m = n*M = [1.474*10^{-7} mol]*[59.93 grams /mol] = 8.81*10^{-6}grams

-----------------------------------------------------------------------------------------

    time t = -[T1/2 / ln2]*ln[N/N0]

             = - [5.3 years / ln2]*ln[1x10-6/1x10-3]

             = 52.8 year

8 0
3 years ago
A 3.50-meter length of wire with a cross-sectional area of 3.14 × 10-6 meter2 is at 20° Celsius. If the wire has a resistance of
maks197457 [2]

Answer:

5.6\times 10^{-8}\ Ohm.m

Explanation:

Resistivity is given by \rho=\frac {AR}{L} where A is cross-sectional area, R is resistance, L is the length and \rho is the reistivity. Substituting 0.0625 for R, 3.14 × 10-6 for A and 3.5 m for L then the resistivity is equivalent to

\rho=\frac {3.14\times 10^{-6}\times 0.0625}{3.5}=5.60714285714285714285714285714285714285\times 10^{-8}\approx 5.6\times 10^{-8}\ Ohm.m

8 0
3 years ago
A 1.50 3 103 - kg car starts from rest and accelerates uniformly to 18.0 m/s in 12.0 s. Assume that air resistance remains const
nexus9112 [7]

Answer:

Explanation:

let force exerted by engine be F.Net force =( F-400)N, applying newton law

     F-400 = 1.5 x 10³x18 =27000 ,

F = 27400 N.

velocity after 12 s  = 0 + 18 x 12 = 216 m/s

Average velocity = (0 + 216 )/2 = 108 m/s

Average power = force x average velocity = 27400 x 108 = 29.6 10⁵ W .⁶

b) At 12 s , velocity = 216 m/s

Instantaneous power = velocity x force = 216 x 27400 = 59.2 x 10⁶ W.

8 0
3 years ago
Read 2 more answers
A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h
Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

v_{1}=\dfrac{3.0\times10^{3}}{3600}

v_{1}=0.833\ m/s

The force F₁is constant acceleration is also a constant.

F_{1}=ma_{1}

We need to calculate the acceleration

Using formula of acceleration

a_{1}=\dfrac{v}{t}

a_{1}=\dfrac{0.833}{10}

a_{1}=0.083\ m/s^2

Similarly,

F_{2}=ma_{2}

For total force,

F_{3}=F_{2}+F_{1}

ma_{3}=ma_{2}+ma_{1}

The speed of second tugboat is

v=\dfrac{11\times10^{3}}{3600}

v=3.05\ m/s

We need to calculate total acceleration

a_{3}=\dfrac{v}{t}

a_{3}=\dfrac{3.05}{10}

a_{3}=0.305\ m/s^2

We need to calculate the acceleration a₂

0.305=a_{2}+0.083

a_{2}=0.305-0.083

a_{2}=0.222\ m/s^2

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}

\dfrac{F_{1}}{F_{2}}=3.7

F_{1}=3.7F_{2}

Hence, The magnitude of F₁ is 3.7 times of F₂

3 0
3 years ago
Are any rocks that form from another rock as a result of change in blank or blank
charle [14.2K]
The correct answer is rock cycle
3 0
3 years ago
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