Answer: 1
an object positioned at some height in a gravitational field
Explanation:
Gravitational potential energy of an object is the energy stored due to position of the object or position at certain height relative to zero position.
Gravitational potential energy can also be expressed as object position at some height above or below zero position in a gravitational field
I think 1 and 2 make sense. But 1 make more sense than 2
The problem ask to calculate the bullet's flight time and the bullet's speed as it left the barrel. So base on the problem, the answer would be that the flight time is 0.076 seconds and the speed of the bullet is 657.9 m/s. I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications.
Straight upward
the ball moves in the forward direction with your walking speed at all times. If you want the ball to land in your hand when it comes back down, you should toss the ball straight upward.
<h3>What is Projectile motion ?</h3>
Projectile motion is the motion of an object thrown (projected) into the air.
- After the initial force that launches the object, it only experiences the force of gravity. The object is called a projectile, and its path is called its trajectory
- A projectile can be a thrown ball, a bullet or a springboard diver ... Except for air resistance, the forward velocity of any projectile is constant and is equal to the initial velocity when it was released.
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This question involves the concepts of density, volume, and mass.
The approximate diameter of a magnesium atom is "3.55 x 10⁻¹⁰ m".
<h3>STEP 1 (FINDING MASS OF INDIVIDUAL ATOM)</h3>
It is given that:
Mass of one mole = 24 grams
Mass of 6 x 10²³ atoms = 24 grams
Mass of 1 atom =
= 4 x 10⁻²³ grams
<h3>STEP 2 (FINDING VOLUME OF A SINGLE ATOM)</h3>

where,
= density = 1.7 grams/cm³- m = mass of single atom = 4 x 10⁻²³ grams
- V = volume of single atom = ?
Therefore,

V = 2.35 x 10⁻²³ cm³
<h3>STEP 3 (FINDING DIAMETER OF ATOM)</h3>
The atom is in a spherical shape. Hence, its Volume can be given as follows:
![V =\frac{\pi d^3}{6}\\\\d=\sqrt[3]{ \frac{6V}{\pi}}\\\\d=\sqrt[3]{ \frac{6(2.35\ x\ 10^{-23}\ cm^3)}{\pi}}](https://tex.z-dn.net/?f=V%20%3D%5Cfrac%7B%5Cpi%20d%5E3%7D%7B6%7D%5C%5C%5C%5Cd%3D%5Csqrt%5B3%5D%7B%20%5Cfrac%7B6V%7D%7B%5Cpi%7D%7D%5C%5C%5C%5Cd%3D%5Csqrt%5B3%5D%7B%20%5Cfrac%7B6%282.35%5C%20x%5C%2010%5E%7B-23%7D%5C%20cm%5E3%29%7D%7B%5Cpi%7D%7D)
d = 0.355 x 10⁻⁷ cm = 3.55 x 10⁻¹⁰ m
Learn more about density here:
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