A gram molecule<span> of a </span>gas<span> at </span>127<span>°C </span>expands isothermally until its volume<span> is </span>doubled<span>. </span>Find<span> the </span>amount<span>of </span>work done<span> and </span>heat absorbed<span>.</span>
No, energy transformation is occurring in every point of the motion.
In fact, the ball starts from point 1 with maximum kinetic energy and zero potential energy (taking the hand of the boy as reference level). The kinetic energy converts into gravitational potential energy as it goes higher: in point 2, part of the kinetic energy has converted into potential energy (because the velocity has decreased, while the height has increased), and then when the ball reaches point 3 all the kinetic energy has converted into potential energy (because now the velocity is zero, while the height is maximum). As the ball descends (point 4), the velocity starts to increase again, therefore the kinetic energy increases and the potential energy decreases (because the height is deacreasing now).
Summarizing, energy transformation is occuring in every point of the motion.
Answer:
exercise can lower levels
Answer:
a. Approximate umber of atoms = 10¹¹
b. number of atoms = 50 atoms thick
Note: The question is missing some parts. The complete question is as follows; Calculate the approximate number of atoms in a bacterium. Assume that the average mass of an atom in the bacterium is ten times the mass of a hydrogen atom. (Hint: The mass of a hydrogen atom is on the order of 1×10⁻²⁷ kg and the mass of a bacterium is on the order of 1×10⁻¹⁵ kg)
Explanation:
a. Approximate umber of atoms = mass of bacterium / 10 * mass of hydrogen
number of atoms = 10⁻⁵/ 10 × 10⁻²⁷
number of atoms = 10¹¹ atoms
b. The cell membrane is 10-8 m thick while the hydrogen atom has a diameter of 10-10 m.
The number of atoms in the cell membrane = d1 / 2 × d2
where d1 = diameter of cell membrane; d2 = diameter of hydrogen atom
The number of atoms in the cell membrane = 10⁻ / 2 × 10⁻¹⁰ = 50 atoms
I don't know I'm just a troll.
