3. A particle of charge +7.5 µC is released from rest at the point x = 60 cm on an x-axis. The particle begins to move due to th
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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
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:
d = 0.355 x 10⁻⁷ cm = 3.55 x 10⁻¹⁰ m
Learn more about density here:
Answer:
a) v = 16.57 m / s, b) a = 19.6 m / s², d) N = 1.76 10³ N, N / W = 3
Explanation:
This exercise looks interesting, but I think you have some problem with the writing, the questions seem a bit disconnected from the initial text.
Let's answer the questions.
a) For this part we can use energy considerations.
Starting point. The upper part of the trajectory indicates that the arm is horizontally
Em₀ = U = m g h
in this case h = r
Final point. For lower of the trajectory
Em_f = K = ½ m v²
as they indicate that there is no friction
Em₀ = em_f
mgh = ½ m v²
v =
let's calculate
v =
v = 16.57 m / s
b) the centripetal acceleration has the formula
a = v² / r
a = 16.57² / 14.0
a = 19.6 m / s²
c) see attached where the diagram is
where N is the normal and w the weight
d) let's use Newton's second law
N-W = m a
N - mg = m ar
N = m (g + a)
let's calculate
N = 60.0 (9.8 + 19.6)
N = 1.76 10³ N
the relationship with weight is
N / W = 1.76 10³/( 60 9.8)
N / W = 3
normal is three times greater than body weight
e) the answer is reasonable since by Newton's first law the body must continue in a straight line, therefore to change its trajectory a force must be applied to deflect it
