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2 years ago

Which of the following decreases with increasing atomic number in Group 2A?

2 answers:

7 0

"Ionization energy" is the one among the following choices given in the question that <span>decreases with increasing atomic number in Group 2A. The correct option among all the options that are given in the question is the third option or option "C". I hope that the answer has helped you.</span>

frozen [14]2 years ago

5 0

1. A. It tends to decrease

2. A. a Positive charge

3. D. lose electrons when they form ions

4. A. It is more difficult to remove a second electron from an atom

5. D. Fluorine

6. B. Ionization energy

7. C. ionization energy

You might be interested in

How do you find a controlled variable

adelina 88 [10]

Hello there!

Essentially, a control variable is what is kept the same throughout the experiment, and it is not of primary concern in the experimental outcome. Any change in a control variable in an experiment would invalidate the correlation of dependent variables (DV) to the independent variable (IV), thus skewing the results.

7 0

3 years ago

An athlete kicks a soccer ball that starts at rest so that it leaves their foot with a speed of 10m/s from the top o f a rectang

kirza4 [7]

Answer:

$a=500m/s^2$

Explanation:

We need only to apply the definition of acceleration, which is:

$a=\frac{v_f-v_i}{t_f-t_i}$

In our case the final velocity is $v_f=10m/s$ , the initial velocity is $v_i=0m/s$ since it departs from rest, the final time is $t_f=0.02s$ and the initial time we are considering is $t_i=0s$

So for our values we have:

$a=\frac{10m/s-0m/s}{0.02s-0s}=500m/s^2$

3 0

2 years ago

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0

1 year ago

An object is located 70 cm from a concave mirror with a focal length of 15 cm. What is the image

Stels [109]

(a) The distance of the image formed by the concave mirror is 19.1 cm.

(b) The image formed is diminished and real.

<h3>Image distance </h3>

The distance of the image formed by the concave mirror is calculated as follows;

1/f = 1/v + 1/u

1/v = 1/f - 1/u

1/v = 1/15 - 1/70

1/v = 0.05238

v = 1/0.05238

v = 19.1 cm

The image distance is smaller than object distance, thus the image formed is diminished and real.

Learn more about concave mirror here: brainly.com/question/13164847

#SPJ1

5 0

1 year ago

Valerie is preparing a blood sample for analysis. She wants to separate the solid red blood cells and obtain only the liquid por

liraira [26]

<span>First, she should put the sample in a test tube and place it in a centrifuge. This would cause the red blood cells to move to the bottom because of their higher density. Next, she would be able to decant the plasma and analyze it separately from the red blood cells.</span>

4 0

3 years ago

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