Hii! help asap. i’ll give brainliest thanks!

Ok I have no clue for this one I’m not sure what to make out of this one please help

MatroZZZ [7]

Helium, Neon, and Xenon are all part of the same column on the Periodic Table. Such a column is referred to as a Group, because they have the same number of valence electrons in their outermost shell. Hope this helps!

5 0

3 years ago

HELP PLEASE 20 ;OINTS

77julia77 [94]

I’m pretty sure it’s C.

7 0

3 years ago

With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you releas

xxTIMURxx [149]

Answer:

The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

Let the initial velocity be 'v'.

Given:

Mass of the ball (m) = 130 g = 0.130 kg [ 1 g = 0.001 kg]

Initial height of the ball (h₁) = 1.4 mm = 0.0014 m [ 1 mm = 0.001 m]

Final height of the ball (h₂) = 15 mm = 0.015 m

Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

$\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity$

As it just touches the 15 mm high roof, the final velocity will be zero. So,

$v_f=0\ m/s$ .

Now, the change in kinetic energy is equal to:

$\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2$

Change in gravitational potential energy = Final PE - Initial PE

So,

$\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J$ [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

$0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s$

Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

4 0

3 years ago

Describe how a neutral material becomes attracted to a negatively charged object brought near it.

Naddik [55]

Answer:

Electric force

Explanation:

It’s like static stuff

3 0

2 years ago

A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve

Gemiola [76]

Answer:

(a) v = 5.42m/s

(b) vo = 4.64m/s

(c) a = 2874.28m/s^2

(d) Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

$v=\sqrt{2gh}$ (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

$v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}$

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

$h_{max}=\frac{v_o^2}{2g}$ (2)