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HACTEHA [7]
3 years ago
15

Why does every element except hydrogen need at least one neutron and its nucleus?

Physics
1 answer:
scoray [572]3 years ago
8 0

Answer:

It needs attractive force from the strong nuclear interaction to counter the electrostatic repulsion between the protons.

Explanation:

It has to counter

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The force involved in falling of an apple from a tree is known as? (a) Magnetic force (b) electrostatic force (c) Contact force
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Answer:

gravitational force

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Phosphorus (P) is an element with an atomic number of 15 and an atomic mass of 31. How many neutrons are in an atom of phosphoru
BaLLatris [955]

(31-15 = 16).

Explanation:

the element phosphorus (P) has an atomic number of 15 and a mass number of 31. Therefore, an atom of phosphorus has 15 protons, 15 electrons, and 16 neutrons

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3 years ago
8. The ratio of output power to input power, in percent?A. work. B. efficiency. C. horsepower. D. conductivity.
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3 years ago
Read 2 more answers
On August 10, 1972, a large meteorite skipped across the atmosphere above the western United States and western Canada, much lik
Anon25 [30]

a) 4.62\cdot 10^{14} J

b) 0.110 megatons

c) 8.46 bombs

Explanation:

a)

The energy lost by the meteorite is equal to the difference between its final kinetic energy and its initial kinetic energy:

\Delta K=K_f-K_i

Which can be rewritten as:

\Delta K=\frac{1}{2}mv^2-\frac{1}{2}mu^2

where:

m=3.2\cdot 10^6 kg is the mass of the meteorite

v=0 is the final speed of the meteorite

u=17 km/s = 17,000 m/s is the initial speed of the meteorite

Substituting the values into the equation, we found the loss in energy of the meteorite:

\Delta K=0-\frac{1}{2}(3.2\cdot 10^6)(17000)^2=-4.62\cdot 10^{14} J

So, the energy lost by the meteorite is 4.62\cdot 10^{14} J

b)

The energy equivalent to 1 megaton of TNT is

E_{TNT}=4.2\cdot 10^{15} J

Here the energy lost by the meteorite is

E=4.62\cdot 10^{14} J

Therefore, in order to write the energy lost by the meteorite as a multiple of the energy of 1 megaton of TNT, we have to divide the energy lost by the meteorite by the energy equivalent to 1 TNT; we find:

\frac{E}{E_{TNT}}=\frac{4.62\cdot 10^{14}}{4.2\cdot 10^{15}}=0.110

So, the energy lost by the meteorite corresponds to 0.110 megatons.

c)

The energy of one atomic bomb explosion in Hiroshima is equal to

E'=13 kt (13 kilotons)

which corresponds to

E'=0.013 Mt (0.013 megatons)

Here the energy of the meteorite is equal to

E=0.110 Mt (0.110 megatons)

Therefore, we can find how many Hiroshima bombs are equivalent to teh meteorite impact by using the following rules of three:

\frac{1 bomb}{0.013 Mt}=\frac{x bombs}{0.110 Mt}\\x=\frac{1\cdot 0.110}{0.013}=8.46

So, 8.46 bombs.

5 0
3 years ago
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