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aliya0001 [1]
3 years ago
10

When 23Na is bombarded with protons, the products are 20Ne and A. a neutron B. an alpha particle C. a deuteron D. a gamma ray pa

rticle E. two beta particles
Physics
1 answer:
Vikki [24]3 years ago
5 0

Answer:

The correct option is B. an alpha particle.

Explanation:

When ²³Na is bombarded with protons _{1}^{1}\textrm{H}, ²⁰Ne and one alpha particle _{2}^{4}\textrm{He} is released.

The reaction is as follows:

_{11}^{23}\textrm{Na} + _{1}^{1}\textrm{H} \rightarrow _{10}^{20}\textrm{Ne} + _{2}^{4}\textrm{He}

Therefore, an alpha particle and ²⁰Ne are released when ²³Na is bombarded with protons.

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saw5 [17]
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8 0
3 years ago
You have a wooden bench in your living room. You set your backpack on the bench and lean your hand against the wall. Name two ac
kotykmax [81]

The action-reaction pairs in the given situation are:

  • the backpack and the bench
  • the and and the wall

<h3>What are action-reaction pairs?</h3>

Action-reaction pairs are two forces which are equal but oppositely directed in their line of action.

Action-reaction pairs are according to Newton's third law of motion.

The action-reaction pairs in the given situation are:

  • the backpack and the bench
  • the and and the wall

Learn more about action-reaction pairs at: brainly.com/question/12800382

#SPJ1

3 0
2 years ago
A waist-to-hip ratio of 1.0 is considered healthy.<br><br> True<br> False
laila [671]
false a waist to hip ratio that is healthy would be .80 or less is considered healthy for most individuals hope this helps
7 0
3 years ago
Read 2 more answers
If the copper is drawn into wire whose diameter is 6.50 mm, how many feet of copper can be obtained from the ingot? the density
Kobotan [32]
Assume that an ingot of copper has a mass of 9.1 kg or 9100 g.

The cross-sectional area of the copper wire with diameter of 6.5 mm (or 0.65 cm) is
A = (π/4)*(0.65 cm)² = 0.3318 cm²

The density of copper is given as 8.94 g/cm³.
If the length of copper wire is L cm, then 
(0.3318 cm²)*(L cm)*(8.94 g/cm³) = 9100 g
L = 9100/(0.3318*8.94) = 3.0678 x 10³ cm

Note that
1 cm = 1/2.54 in = 1/2.54 in = 0.3937 in
        = 0.3937/12 = 0.03281 ft

Therefore
L = (3.0678 x 10³ cm)*(0.03281 ft/cm) = 100.65 ft

Answer: 100.65 ft

6 0
4 years ago
Rank the six combinations of electric charges on the basis of the electric force acting on q1. Define forces pointing to the rig
g100num [7]

I have attached the image showing the 6 combination of charges

Answer:

Smallest are: q1 = +1 nc, q2 = +1 nc, q3 = +1 nc and q1 = -1 nc, q2 = -1 nc, q3 = -1nc

Third Largest: q1 = +1 nc, q2 = +1nc, q3 = -1 nc

Second Largest: q1 = +1 nc, q2 = -1 nc, q3 = +1 nc

Largest are: q1 = +1 nc, q2 = -1 nc, q3 = -1 nc and q1 = -1 nc, q2 = +1 nc, q3 = +1 nc

Explanation:

The electrostatic force between 2 charges is; F = k•q1•q2/r²

Where r is the distance between the 2 charges q1 and q2 and k is coulombs constant.

Thus;

F ∝ q1•q2/r²

The net force on the charge q1 is the sum of the forces on charge q2 and q3.

If the two charges are opposite, the force is an attractive force while if the two charges are similar, the force is a repulsive force.

Thus, if the combination has same type of charges, i.e;(+1,+1,+1) or (-1,-1,-1), the force will be very small.

So, now the order of forces from smallest to largest is;

Smallest are: q1 = +1 nc, q2 = +1 nc, q3 = +1 nc and q1 = -1 nc, q2 = -1 nc, q3 = -1nc

Third Largest: q1 = +1 nc, q2 = +1nc, q3 = -1 nc

Second Largest: q1 = +1 nc, q2 = -1 nc, q3 = +1 nc

Largest are: q1 = +1 nc, q2 = -1 nc, q3 = -1 nc and q1 = -1 nc, q2 = +1 nc, q3 = +1 nc

3 0
3 years ago
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