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Ulleksa [173]
3 years ago
8

What is the change in atomic mass number when an atom emits an alpha particle?

Physics
1 answer:
natita [175]3 years ago
6 0

Answer:

Decreases by 4

Explanation:

  • An alpha particle is made of two protons and two neutrons. An alpha decay involves the ejection of an alpha particle from the nucleus.
  • Therefore since an alpha particles is equivalent to a helium nucleus, it involves the loss of 2 protons and 2 neutrons.
  • <em><u>Hence, we can conclude that when a nucleus emits an alpha particle, the mass number or the atomic mass decreases by 4 and the atomic number decreases by 2</u></em><em><u>.</u></em>
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Work occurs when
defon
I believe the answer is the second option.
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3 years ago
A toy car is moving across a table with a velocity of 7.0 m/s and drives off the end. The table is 1.8 m tall. How long will the
KatRina [158]

Answer:

Option D - 0.2 s

Explanation:

We are given;

Initial velocity; u = 7 m/s

Height of table; h = 1.8m

Now,since we want to find the time the car spent in the air, we will simply use one of Newton's equation of motion.

Thus;

h = ut + ½gt²

Plugging in the relevant values, we have;

1.8 = 7t + ½(9.8)t²

4.9t² + 7t - 1.8 = 0

Using quadratic formula to find the roots of the equation gives us;

t = -1.65 or 0.22

We can't have negative t value, thus we will pick the positive one.

So, t = 0.22 s

This is approximately 0.2 s

7 0
3 years ago
What is dark energy?
expeople1 [14]
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4 0
3 years ago
I have a special thermos that has a vacuum chamber between my hot soup and its outside wall. Because my soup is isolated from it
bagirrra123 [75]

Answer:

C. My soup will cool down because of radiation.

Explanation:

Since, there is vacuum between hot soup and its outside wall, then heat can not flow through conduction and convection.

The heat then only flows through radiation.

Therefore, the soup will not cool down because of convection or conduction by because of radiation.

7 0
3 years ago
Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge
NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is V=\frac{1}{4\pi\epsilon} \frac{q}{d}

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as-E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero  between the two charges,but we can find the points situated on the axial  line but  outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining  two charges  where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0
3 years ago
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