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2 years ago

What is the source of the radioactive nuclei present in spent fuel rods?

1 answer:

6 0

The answer is nuclear fission. The source of the radioactive nuclei present in spent fuel rods is nuclear fission. Nuclear fission is the process in which a large nucleus splits into two smaller nuclei with the release of energy.

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Water is formed when two hydrogen atoms bond to an oxygen atom. The hydrogen and the oxygen in this example are different

Goryan [66]

They both are two different elements.SO the answer is Element

8 0

3 years ago

A 1000 kg satellite and a 2000 kg satellite follow exactly the same orbit around the earth. What is the ratio F1/F2 of the gravi

Tamiku [17]

Answer:

the ratio F1/F2 = 1/2

the ratio a1/a2 = 1

Explanation:

The force that both satellites experience is:

F1 = G M_e m1 / r² and

F2 = G M_e m2 / r²

where

m1 is the mass of satellite 1
m2 is the mass of satellite 2
r is the orbital radius
M_e is the mass of Earth

Therefore,

F1/F2 = [G M_e m1 / r²] / [G M_e m2 / r²]

F1/F2 = [G M_e m1 / r²] × [r² / G M_e m2]

F1/F2 = m1/m2

F1/F2 = 1000/2000

F1/F2 = 1/2

The other force that the two satellites experience is the centripetal force. Therefore,

F1c = m1 v² / r and

F2c = m2 v² / r

where

m1 is the mass of satellite 1
m2 is the mass of satellite 2
v is the orbital velocity
r is the orbital velocity

Thus,

a1 = v² / r ⇒ v² = r a1 and

a2 = v² / r ⇒ v² = r a2

Therefore,

F1c = m1 a1 r / r = m1 a1

F2c = m2 a2 r / r = m2 a2

In order for the satellites to stay in orbit, the gravitational force must equal the centripetal force. Thus,

F1 = F1c

G M_e m1 / r² = m1 a1

a1 = G M_e / r²

also

a2 = G M_e / r²

Thus,

a1/a2 = [G M_e / r²] / [G M_e / r²]

a1/a2 = 1

4 0

3 years ago

How is energy transferred during the water cycle? Question 1 options: Water gains energy during evaporation and releases it duri

Margaret [11]

Answer:

Water gains energy during evaporation and releases it during condensation in the atmosphere

Explanation:

In the water cycle, heat energy is gained or lost by water as it undergoes various processes in the cycle.

In evaporation, water molecules gains energy because the molecules of water vibrate faster and become more energetic. Hence they are able to escape into the atmosphere from the surface of the liquid.

In condensation, the molecules of gaseous water looses energy and becomes liquid.

Hence, water gains energy during evaporation and releases it during condensation in the atmosphere.

8 0

2 years ago

Read 2 more answers

A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur

muminat

Answer:

a) E = 0

b) $E = \dfrac{k_e \cdot q}{ r^2 }$

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

From which we have;

$E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0$

E = 0/A = 0

E = 0

b) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

$E \cdot A = \dfrac{+q }{\varepsilon _{0}}$

$E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}$

By Gauss theorem, we have;

$E\oint dS = \dfrac{q}{\varepsilon _{0}}$

Therefore, we get;

$E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}$

The electrical field outside the spherical shell

$E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }$