Ask question

Ask question

All categories

3 years ago

15

A fission reaction possible is that certain atoms are

1 answer:

Elden [556K]3 years ago

8 0

The process of splitting a nucleus<span> is called </span>nuclear fission<span>. Uranium or plutonium isotopes are normally used as the fuel in </span>nuclear<span> reactors because their </span>nuclei<span> are relatively large and easy to split. For </span>fission<span> to </span>happen<span>, the uranium-235 or plutonium-239 </span>nucleus<span> must first absorb a neutron.</span>

You might be interested in

Lithium has a density of 0.5 g/cm3. What would be the mass of a 20 cm3 sample of lithium?

MrMuchimi

Answer:

1 gm

Explanation:

.05 g/cm^3 * 20 cm^3 = 1 gm

6 0

2 years ago

From Kepler's third law, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to:

bazaltina [42]

Answer:

The correct option is (B).

Explanation:

The Kepler's third law of motion gives the relationship between the orbital time period and the distance from the semi major axis such that,

$T^2\propto a^3\\\\T^2=ka^3$

It is mentioned that, an asteroid with an orbital period of 8 years. So,

$(8)^2=ka^3\\\\64=ka^3\\\\a=(64)^{\dfrac{1}{3}}\\\\a=4\ AU$

So, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 4 astronomical units.

7 0

3 years ago

A mechanic is trying to loosen a nut with a wrench, but it is stuck. What could the mechanic do to help loosen the nut? (Hint: T

SIZIF [17.4K]

C. Use a longer wrench, this is because it would create a greater turning moment as moment=force*distance

5 0

3 years ago

A boy drops a coin down a well that is 225 m deep. How long does it take the coin to hit the bottom of the well? (Please state y

Alexxandr [17]

I had the same question. I'm pretty sure it's 23 seconds. :)

7 0

3 years ago

If 2050 J of heat are added to a 150 g object its temperature increases by 15°C.

Darina [25.2K]

When an object gets heated by a temperature ΔT energy needed, E = mcΔT

Here energy is given E = 2050 J

Mass of object = 150 g

Change in temperature ΔT = 15 $^0C$ = 15 K

a) Heat capacity of an object equal to the ratio of the heat added to (or removed from) an object to the resulting temperature change.

So heat capacity = E/ΔT = 2050/15 = 136.67 J/K

b) We have E = mcΔT

c = $\frac{2050}{150*10^{-3}*15} = 911.11 J/kgK$

So object's specific heat = 911.11 J/kgK

5 0

3 years ago