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3 years ago

A neutral atom as the same number of protons and electrons.

1 answer:

6 0

That is correct. Or so I believe. Either more or less than the other on the amount of protons and electrons, you can get either an unstable or a stable atom of an element.

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Antonina throws a coin straight up from a height of

vichka [17]

Answer:

$s=vt-\frac{1}{2}gt^2$

Explanation:

We could use the following suvat equation:

$s=vt-\frac{1}{2}gt^2$

where

s is the vertical displacement of the coin

v is its final velocity, when it hits the water

t is the time

g is the acceleration of gravity

Taking upward as positive direction, in this problem we have:

s = -1.2 m

$g=-9.8 m/s^2$

And the coin reaches the water when

t = 1.3 s

Substituting these data, we can find v:

$v=\frac{s}{t}+\frac{1}{2}gt=-\frac{1.2}{1.3}+\frac{1}{2}(-9.8)(1.3)=-7.3 m/s$

where the negative sign means the direction is downward.

5 0

3 years ago

What type of energy does mechanical energy turn into when friction is acting

umka2103 [35]

Thermal energy is the answer

Hope that helps

6 0

3 years ago

Design your own plan for a scientific experiment. You do not have to conduct your experiment.

olganol [36]

you could use one of your scientific experiments you did over the past few years.

You could make an solinoid.

Hope this helps! Lol

8 0

3 years ago

A 0.69 kg rubber ball has a speed of 1.4 m/s at point a and kinetic energy 7.5 j at point

NikAS [45]

Dddddddddddddddwwwwwwdwwwww

7 0

3 years ago

A glider moves along an air track with constant acceleration a. It is projected from the start of the track (x = 0 m) with an in

Pie

Answer:

vo = 0.175m/s

a = -0.040625 m/s^2

Explanation:

To solve this problem, you will need to use the equations for constant acceleration motion:

$x = \frac{1}{2}at^2 +v_ot+x_o \\v_f^2 - v_o^2 = 2(x-x_o)a$

In the first equation you relate final position with the time elapsed, in the second one, you relate final velocity at any given position. In both equations, you will have both the acceleration a and the initial velocity vo as variables. We can simplify with the information we have:

1. $x = \frac{1}{2}at^2 +v_ot+x_o\\0.1m = \frac{1}{2}a(8s)^2 +v_o(8s)+0m \\0.1 = 32a + 8v_o$

2. $v_f^2 - v_o^2 = 2(x-x_o)a\\(-0.15m/s)^2 - v_o^2 = 2(0.1m-0m)a\\0.0225 - v_o^2 = 0.2a\\a = \frac{0.0225 - v_o^2}{0.2} = 0.1125 - 5v_o^2$

Replacing in the first equation:

$0.1 = 32(0.1125 - 5v_o^2) + 8v_o\\0.1 = 3.6 - 160v_o^2 + 8v_o\\160v_o^2 - 8v_o - 3.5 = 0$

$v_0 = \frac{-(-8) +- \sqrt{(-8)^2 - 4(160)(-3.5)}}{2(160)} \\ v_o = 0.175 m/s | -0.125 m/s$

But as you are told that the ball was projected om the air track, it only makes sense for the velocity to be positive, otherwise it would have started moving outside the air track, so the real solution is 0.175m/s. Then, the acceleration would be:

$a = 0.1125 - 5v_o^2\\a = -0.040625 m/s^2$

3 0

4 years ago