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:
brainly.com/question/952755
Answer:
h~=371.26m
Explanation:
when an object falls we use the equations of accelerated motion. There is only one that gives distance.
Since we have no initial velocity (started from rest) we can get rid of the (ut) term
where a we substitute g (gravitational acceleration, constant for given heights and almost 9.81m/s^2).
Answer:
1) t = 3.45 s, 2) x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s
,
5) θ = -40.2º
Explanation:
This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.
1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff
y = y₀ + 
t - ½ g t²
When leaving the cliff the speed is horizontal v_{oy}= 0 and at the bottom of the cliff y = 0
0 = y₀ - ½ g t2
t = √ 2y₀ / g
t = √ (2 60 / 9.8)
t = 3.45 s
2) The horizontal distance traveled
x = v₀ₓ t
x = 40 3.45
x = 138 m
3) The vertical velocity at the point of impact
v_{y} = I go - g t
v_{y} = 0 - 9.8 3.45
v_{y} = -33.81 m /s
the negative sign indicates that the speed is down
4) the resulting velocity at this point
v = √ (vₓ² + v_{y}²)
v = √ (40² + 33.8²)
v = 52.37 m / s
5) angle of impact
tan θ = v_{y} / vx
θ = tan⁻¹ v_{y} / vx
θ = tan⁻¹ (-33.81 / 40)
θ = -40.2º
6) sin (-40.2) = -0.6455
7) tan (-40.2) = -0.845
8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis
Answer:
... If your arrows are too lightly or heavily spined for your bow, the “archer's paradox” resulting in poor arrow flight and loss of accuracy.
