The range of the piece of paper is C) 1.4 m
Explanation:
The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:
- A uniform motion along the horizontal direction, with constant velocity
- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity,
)
From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

where
u is the initial velocity
is the angle of projection
g is the acceleration of gravity
For the piece of paper in this problem,
u = 4.3 m/s

Substituting,

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This type of listening response is called back-channel signal. This allows the speaker to know that the listener is attentive or willing to engage a conversation between them. It is shown through short utterances, facial expressions, head nods and others.
<span>Here the force that is applied between the electron and proton is centripetal, so equate the two forces to determine the velocity.
We know charge of the electron which for both Q1 and Q2, e = 1.60 x 10^-19 C
The Coulombs Constant k = 9.0 x 10^9
Radius r = 0.053 x 10^-9m = 5.3 x 10^-11 m
Mass of the Electron = 9.11 x 10^-31
F = k x Q1 x Q2 / r^2 = m x v^2 / r(centripetal force)
ke^2 / r^2 = m x v^2 / r => v^2 = ke^2 / m x r
v^2 = ((1.60 x 10^-19)^2 x 9.0 x 10^9) / (9.11 x 10^-31 x 5.3 x 10^-11 )
v^2 = 4.77 x 10^12 = 2.18 x 10^6 m/s
Since one orbit is the distance,
one orbit = circumference = 2 x pi x r; distance s = v x t.
v x t = 2 x pi x r => t = (2 x 3.14 x 5.3 x 10^-11) / (2.18 x 10^6)
t = 33.3 x 10^-11 / 2.18 x 10^6 = 15.27 x 10^-17 s
Revolutions per sec = 1 / t = 1 / 15.27 x 10^-17 = 6.54 x 10^15 Hz</span>
Explanation:
The structural diversity of carbon-based molecules is determined by following properties:
1. the ability of those bonds to rotate freely,
2.the ability of carbon to form four covalent bonds,
3.the orientation of those bonds in the form of a tetrahedron.
Answer:
assume nitrogen is an ideal gas with cv=5R/2
assume argon is an ideal gas with cv=3R/2
n1=4moles
n2=2.5 moles
t1=75°C <em>in kelvin</em> t1=75+273
t1=348K
T2=130°C <em>in kelvin</em> t2=130+273
t2=403K
u=пCVΔT
U(N₂)+U(Argon)=0
<em>putting values:</em>
=>4x(5R/2)x(Tfinal-348)=2.5x(3R/2)x(T final-403)
<em>by simplifying:</em>
Tfinal=363K