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aliya0001 [1]
3 years ago
13

Which statement is correct? When a positively charged atom looses an electron to a positively charged atom, two neutral atoms ar

e created. When a neutral atom looses an electron to another neutral atom, two charged atoms are created. When a neutral atom looses an electron to a positively charged atom, both atoms become neutral. When charged atoms lose electrons they will become positively charged atoms.
Physics
2 answers:
Setler79 [48]3 years ago
7 0

Answer:

When a neutral atom looses an electron to another neutral atom, two charged atoms are created.

Explanation:

Initially, we have two neutral atoms: they both have electrical charge equal to zero.

If one of the two neutral atom looses an electron, it becomes positively charge (charge: +1), because the electron carries a charge of -1. As a result, the other atom which accepts the electron becomes negatively charged (charge: -1). Therefore, in the end we will have two charged atoms.

vazorg [7]3 years ago
3 0
I believe the answer is "When a neutral atom looses an electron to another neutral atom, two charged atoms are created."
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Please Help!!!! I WILL GIVE BRAINLIEST!!!!!!!!!!!!!
Bas_tet [7]

Given info

d = 0.000250 meters = distance between slits

L = 302 cm = 0.302 meters = distance from slits to screen

\theta_8 = 1.12^{\circ} = angle to 8th max (note how m = 8 since we're comparing this to the form \theta_m)

x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters} (n = 5 as we're dealing with the 5th minimum )

---------------

Method 1

d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}

Make sure your calculator is in degree mode.

-----------------

Method 2

\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\

-----------------

Method 3

\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\

There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.

7 0
3 years ago
Prefix and suffix for hydrology
Tju [1.3M]

Prefix: Hydro

Suffix: Logy

I hope I helped you! <3

8 0
3 years ago
A 20 kg object is dropped from a very tall building. What is the weight of this objects? After 5 seconds, how has the object fal
prohojiy [21]

1. What is the weight of this objects?

Weight is simply the product of mass and gravitational acceleration. Therefore the weight is:

w = 20 kg * 9.81 m/s^2

w = 196.2 kg m/s^2 = 196.2 N

 

2. After 5 seconds, how has the object fallen and what is its speed at this instant?

We can use the formula:

<span>y = v0 t  + 0.5 g t^2</span>

v = v0 + g t

where v0 = 0 since the object starts from rest, y is the distance it fell, t is time

y = 0 + 0.5 * 9.81 * 5^2 = 122.625 m

<span>v = 0 + 9.81 * 5 = 49.05 m/s</span>

8 0
3 years ago
Jack travelled 360 km at an average speed of 80 km/h. Elaine
diamong [38]

Answer:

Average speed of Elain = 60 km/h

Explanation:

Total Distance covered by Jack = 360km

Average Speed of Jack = 80 km/h

Time taken by Jack to complete his journey = Distance / Average speed = 360 km / 80 km/h

Time taken by Jack to complete his journey = 4.5 hours

As it is given the both Jack and Elain travelled the same amount of distance:

Total distance travelled by Elain = 360 km

It is given that Elain took 1.5 hourse more than Jack to cover the distance, so Time taken by Elain to cover the distance is = 4.5 hours + 1.5 hours = 6 hours

Average speed of Elain = Distance/ time = 360 km / 6 hours

Average speed of Elain = 60 km/h

3 0
3 years ago
Light propagate faster through medium “a” than medium “b”
dangina [55]

1) Medium "b" has more optical density

2) Light must hit the interface between the two mediums perpendicularly

Explanation:

1)

Refraction occurs when light propagates from a medium into a second medium.

The optical density of a medium is given by its index of refraction, which is defined as:

n=\frac{c}{v}

where

c is the speed of light in a vacuum

v is the speed of light in a medium

Higher index of refraction means higher optical density, and light propagater slower into a medium with higher optical density.

In this problem, light propagates faster through medium "a" than medium "b": this means that medium "a" has lower refractive index of medium "b", and so "b" has more optical density.

2)

We can answer this part by referring to Snell's law, which gives the relationship between the direction of the incident ray and of the refracted ray when light passes through the interface between two media:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n_1, n_2 are the index of refraction of the two mediums

\theta_1, \theta_2 are the angle of incidence and of refraction (the angle that light makes with the normal to the surface in medium 1 and medium 2)

Here we want the direction of propagation of the light ray not to change: this means that it must be

sin \theta_1 = sin \theta_2 (1)

However, here we have two mediums "a" and "b" with different index of refraction, so

n_1\neq n_2

Therefore the only angle that can satisfy eq.(1) is

\theta_1 = \theta_2 = 0

So, the light must hit the surface perpendicular to the interface between the two mediums.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

3 0
3 years ago
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