Mass of an electron = 9.110 x 10⁻³¹ kg.
Mass of a proton = 1.6727 x 10⁻²⁷ kg
∴ mass of a proton/mass of an electron = 1.6727 x 10⁻²⁷ kg/9.110 x 10⁻³¹ kg.
~1836
∴ mass of a proton = 1836 x mass of an electron.
∴ mass of an electron is insignificant to the mass of an atom.
∴mass of an atom = mass of protons + mass of neutrons
Answer:
"Both students are correct. When light is traveling through small openings, solids, liquids, gases, or a vacuum, it behaves like a wave. When it interacts with matter, it acts like a stream of particles."
Explanation:
just got it right
Answer:
1.23×10⁸ m
Explanation:
Acceleration due to gravity is:
a = GM / r²
where G is the universal gravitational constant,
M is the mass of the planet,
and r is the distance from the center of the planet to the object.
When the object is on the surface of the Earth, a = g and r = R.
g = GM / R²
When the object is at height i above the surface, a = 1/410 g and r = i + R.
1/410 g = GM / (i + R)²
Divide the first equation by the second:
g / (1/410 g) = (GM / R²) / (GM / (i + R)²)
410 = (i + R)² / R²
410 R² = (i + R)²
410 R² = i² + 2iR + R²
0 = i² + 2iR − 409R²
Solve with quadratic formula:
i = [ -2R ± √((2R)² − 4(1)(-409R²)) ] / 2(1)
i = [ -2R ± √(1640R²) ] / 2
i = (-2R ± 2R√410) / 2
i = -R ± R√410
i = (-1 ± √410) R
Since i > 0:
i = (-1 + √410) R
R = 6.37×10⁶ m:
i ≈ 1.23×10⁸ m
ANSWER
EXPLANATION
Parameters given:
Mass of the student, M = 70 kg
Mass of the textbook, m = 1 kg
Distance, r = 1 m
To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:
where G = gravitational constant
Therefore, the gravitational force acting between the student and the textbook is:
That is the answer.
