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mestny [16]
3 years ago
15

How would the composition of an atom change if both the atomic number and mass number each increase by one? A) The atom would ha

ve 1 more neutron. B) The atom would have one more proton and electron. C) The atom would have one more proton and one more neutron. D) The atom would have one more proton, one more neutron and one more electron.
Physics
1 answer:
weqwewe [10]3 years ago
8 0
Choice-B is the correct one. 

-- The atomic number is the number of protons in the nucleus.
-- Each proton in the nucleus is usually matched by one electron in the 'cloud'.
-- The addition of a proton OR a neutron increases the mass number by 1 .
-- Electrons have such small mass that they don't figure into the atomic mass at all.
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What is the technical name for the animal kingdom
enyata [817]
The community off animals
6 0
3 years ago
food product with 10 kg mass is being transported to thesurface of the moon,where the acceleration due to gravity is1.624 m/s2;
larisa [96]

Answer:

In SI units 98.1 N, 16.24 N

English units 22.053861 lbf, 3.6509144 lbf

Explanation:

g = Acceleration due to gravity

m = Mass = 10 kg

Weight on Earth

W=mg\\\Rightarrow W=10\times 9.81\\\Rightarrow W=98.1\ N

Converting to lbf

98.1\times 0.22481=22.053861\ lbf

On Moon

W=10\times 1.624\\\Rightarrow W=16.24\ N

Converting to lbf

16.24\times 0.22481=3.6509144\ lbf

In SI units 98.1 N, 16.24 N

English units 22.053861 lbf, 3.6509144 lbf

5 0
3 years ago
Why are different constellations<br> of stars seen during different<br> seasons?
slamgirl [31]
Actually, they're not.  There's a group of stars and constellations arranged
around the pole of the sky that's visible at any time of any dark, clear night,
all year around.  And any star or constellation in the rest of the sky is visible
for roughly 11 out of every 12 months ... at SOME time of the night. 

Constellations appear to change drastically from one season to the next,
and even from one month to the next, only if you do your stargazing around
the same time every night.

Why does the night sky change at various times of the year ?  Here's how to
think about it:

The Earth spins once a day. You spin along with the Earth, and your clock is
built to follow the sun . "Noon" is the time when the sun is directly over your
head, and "Midnight" is the time when the sun is directly beneath your feet.

Let's say that you go out and look at the stars tonight at midnight, when you're
facing directly away from the sun.

In 6 months from now, when you and the Earth are halfway around on the other
side of the sun, where are those same stars ?  Now they're straight in the
direction of the sun.  So they're directly overhead at Noon, not at Midnight.

THAT's why stars and constellations appear to be in a different part of the sky,
at the same time of night on different dates.
5 0
3 years ago
Read 2 more answers
Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu
vlada-n [284]

Answer:

\frac{1}{10}M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒I_{1} \times W_{1} =I_{2} \times W_{2}

  • I_{1} is the moment of interia of earth before impact
  • W_{1} is the angular velocity of earth about an axis passing through the center of earth before impact
  • I_{2} is moment of interia of earth and asteroid system
  • W_{2} is the angular velocity of earth and asteroid system about the same axis

let  W_{1}=W

since \text{Time period of rotation}∝\frac{1}{\text{Angular velocity}}

⇒ if time period is to increase by 25%, which is \frac{5}{4} times, the angular velocity decreases 25% which is \frac{4}{5}  times

therefore W_{1} = \frac{4}{5} \times W_{1}

I_{1}=\frac{2}{5} \times M\times R^{2}(moment of inertia of solid sphere)

where M is mass of earth

           R is radius of earth

I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where M_{1} is mass of asteroid

⇒ \frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})

\frac{1}{2} \times M\times R^{2}= (\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})

M_{1}\times R^{2}= \frac{1}{10} \times M\times R^{2}

⇒M_{1}=}\frac{1}{10} \times M

3 0
3 years ago
Merton’s strain theory would have the most trouble explaining which crime?
Ronch [10]
Strain theory would least apply to assault. Strain theory represents the result of society pressuring individuals to conform or to achieve accepted goals. When these goals, such as not making enough money or not being well known, fraud, robbery, and burglaries will take place to counteract it. Assault, on the other hand, is used as a bridge into crimes such as fraud, robbery, or burglary.
5 0
3 years ago
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