Answer:
3 micro coulombs = 3 * 10E-6 coulombs charge of aluminum
13 * 10E15 * 1.6 * 10E-19 = 2.08 E-3 Coulombs - charge of atoms in Al
3 * 10E-6 / 2.08 * E-3 = 1.44 * E-3 = .00144 = .144 %
.00144 of the original electrons would have to be lost
Answer: Hello there
Explanation:
The colored part of the eye which helps regulate the amount of light entering the eye. When there is bright light, the iris closes the pupil to let in less light. And when there is low light, the iris opens up the pupil to let in more light. Focuses light rays onto the retina.
Answer:
The car must travel 51.34 m for its speed to reach 2.7 m/s
Explanation:
Mass of car = 2.9 x 10³ kg
Forward force = 1148 N
Resistive force = 941 N
Total force = 1148 - 941 = 207 N
We know
Force = Mass x Acceleration
207 = 2.9 x 10³ x Acceleration
Acceleration = 0.071 m/s²
Now we have equation of motion, v² = u² + 2as
Initial velocity, u = 0 m/s
Final velocity, v = 2.7 m/s
Acceleration, a = 0.071 m/s²
Substituting
v² = u² + 2as
2.7² = 0² + 2 x 0.071 x s
s = 51.34 m
The car must travel 51.34 m for its speed to reach 2.7 m/s
Answer:
Star A would be brighter than Star B
Explanation:
The apparent brightness of a star as perceived on Earth is dependent on its temperature, size, luminosity and distance from the Earth. Apparent brightness is the visible brightness to the eye at the surface of the Earth, while luminosity is the true brightness at the surface of the star.
A hotter star will radiate more energy per second per meter square of surface area. A larger star will have a greater surface area for radiation of energy, thus increasing the luminosity. For two identical stars, the difference in apparent brightness will be dependent on their distances from Earth.
Brightness and distance from earth have an inverse square relationship.
∝
Assuming the star is a point source of radiation, as distance from the source is increased, the radiation is distributed over a surface proportional to the distance form the source. As distance is further increased, the radiation is distributed over a larger surface area reducing the effective luminosity.
If one star (Star B) is twice as far from the earth as the first (Star A), the brightness of Star B will be of Star A.
Thus, Star B will appear to be a quarter of the brightness of Star A. Or, Star A will appear to be 4 times as bright as Star B.