Answer:

Option A is correct
Explanation:
Let's check the options:
A: The electron has a negative charge and is found outside of the nucleus.
Yeah! It's TRUE . An electron is a <u>negatively</u><u> </u><u>charged</u><u> </u><u>particle</u><u> </u><u>and</u><u> </u><u>is</u><u> </u><u>located</u><u> </u><u>outside</u><u> </u><u>the</u><u> </u><u>nucleus</u><u> </u><u>of</u><u> </u><u>an</u><u> </u><u>atom.</u>
B : The neutron has a negative charge and is found in the nucleus.
No! It's FALSE . A neutron carries <u>no </u><u>charge</u>. i.e it is a neutral particle and found inside the nucleus.
C : The proton has no charge and is found in the nucleus.
No! It's FALSE.A proton is a <u>positively</u><u> </u><u>charged</u><u> </u><u>particle</u> present inside nucleus of an atom.
D : The neutron has no charge and is found outside of the nucleus.
I agree that the neutron has no charge. But it is found <u>inside</u> the nucleus not outside . So, this statement is FALSE .
Hence, we found our answer! :D
A. The electron has a negative charge and is found outside of the nucleus is the correct statement about an atom.
Hope I helped!
Best regards! :D
~
Answer:
6.1 km
Explanation:
Given that a plane travels 4.0 km at an angle of 25◦ to the ground, then changes direction and travels 10 km at an angle of 16◦ to the ground. What is the magnitude of the plane's total displacement? Answer in units of km
The magnitude of the total displacement D can be calculated by using cosine formula
Ø = 25 - 16 = 9 degree
D^2 = 4^2 + 10^2 - 2 × 4 × 10 × cos 9
D^2 = 16 + 100 - 80cos9
D^2 = 116 - 79.02
D = sqrt( 36.98)
D = 6.1 m
Therefore, the magnitude of the plane's total displacement is 6.1 km
3) Earth is about 150 million km from the Sun, and the apparent brightness of the Sun in our sky is about 1,300 watts per square meter. Determine the apparent brightness we would measure for the Sun if we were located five times Earth's distance from the Sun. Answer: The Sun would appear 1/25 times as bright.