<span>If you want the Lewis diagram of arsenic, it is a single element, and not much of a "structure". Simply place five dots on the four sides of an imaginary square around the symbol As. Two of the dots are paired, three are unpaired.
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By definition of average acceleration,
<em>a</em> = (20 m/s - 33.1 m/s) / (4.7 s) ≈ -2.78 m/s²
Vertically, the car is in equilibrium, so the net force is equal to the friction force in the direction opposite the car's motion:
∑ <em>F</em> = (1502.7 kg) (-2.78 m/s²) ≈ -4188.38 N ≈ -4200 N
If you just want the magnitude, drop the negative sign.
Answer:
μ = 0.0315
Explanation:
Since the car moves on a horizontal surface, if we sum forces equal to zero on the Y-axis, we can determine the value of the normal force exerted by the ground on the vehicle. This force is equal to the weight of the cart (product of its mass by gravity)
N = m*g (1)
The friction force is equal to the product of the normal force by the coefficient of friction.
F = μ*N (2)
This way replacing 1 in 2, we have:
F = μ*m*g (2)
Using the theorem of work and energy, which tells us that the sum of the potential and kinetic energies and the work done on a body is equal to the final kinetic energy of the body. We can determine an equation that relates the frictional force to the initial speed of the carriage, so we will determine the coefficient of friction.

where:
vf = final velocity = 0
vi = initial velocity = 85 [km/h] = 23.61 [m/s]
d = displacement = 900 [m]
F = friction force [N]
The final velocity is zero since when the vehicle has traveled 900 meters its velocity is zero.
Now replacing:
(1/2)*m*(23.61)^2 = μ*m*g*d
0.5*(23.61)^2 = μ*9,81*900
μ = 0.0315
Answer:
r2 = 2.401557 cm
distance = 0.10 cm
Explanation:
given data
radius = 2.50 cm
density = 15.0 nC/m
voltmeter read = 175
solution
we know here potential difference that is express as
ΔV =
...........1
so here
as here
is linear charge density
r2 = r1 ×
r2 = 2.40 ×
r2 = 2.401557 cm
and
here distance above surface will be
distance = r1 - r2
distance = 2.50 - 2.40
distance = 0.10 cm