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mina [271]
3 years ago
12

How many neutrons are present in an atom of Argon-38

Physics
2 answers:
Katen [24]3 years ago
6 0

Answer:

since 22 isnt an option id pick d

Explanation:

Free_Kalibri [48]3 years ago
3 0

Answer:

22

Hope this helps :)

Have a great day !

5INGH

Explanation:

18 protons so 18 electrons

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Help with this please!
DochEvi [55]

Answer:

I have no clue.

Explanation:

3 0
3 years ago
Particles 1 and 2 of charge q1 = q2 = +3.20 × 10−19 C are on a y axis at distance d = 17.0 cm from the origin. Particle 3 of cha
mel-nik [20]

Answer:

(a) 0.17 m

(b) 5.003 m

(c) 6.38 × 10^{-26} N

(d) 7.37 ×10^{-29} N

Explanation:

(a) The minimum value of x will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

h^{2} = b^{2} + p^{2}

h^{2} = 5^{2} + 0.17^{2}   \\h = \sqrt{} 25.03\\h= 5.002 m

<em>Hence, the maximum distance is 5.002 m</em>

(c) For minimum magnitude we use the minimum distance calculated in (a)

Minimum Distance = 0.17 m

For electrostatic force=     F=\frac{kq1q2}{x^{2} }

F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19}  }{0.17^{2} }

F= 6.38×10^{-26} N

(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F =  \frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19}   }{5.002^{2} } }

F= 7.37×10^{-29 N

4 0
3 years ago
Which is true about density?
Anton [14]

Answer:

Density is independent of the

amount of the sample. (D)

Explanation:

Because density is an intrinsic property of matter.

hope it helps!

3 0
3 years ago
Read 2 more answers
Paula has walked in a straight line, 30.5° north of west, for 1650 meters. How far south and east should she walks to return to
olchik [2.2K]
The answer is A.
Sy = 1650 x sin30.5 = 837.4 m toward south
Sx = 1650 x cos30.5 = 1421.7 m toward east
8 0
3 years ago
A truck on the freeway originally moving at 6.6 meters/second accelerates uniformly with acceleration a = 2.8 meters/second2 for
Sonja [21]

Answer:

139.514 metres

Explanation:

Initial velocity of the truck = 6.6 m/s

Acceleration of the truck = 2.8 m/s^2

Time interval = 7.9 s

Therefore we use the formula,

s = ut + 1/2 at^2

*where s(the distance travelled)...u(the initial velocity)...t(the time period)

; s = 6.6(7.9) + 1/2 (2.8)(7.9)^2

; s = 52.14 + 87.374

The distance moved by the truck = 139.514m

8 0
3 years ago
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