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Alexxandr [17]
3 years ago
6

Classify Use periodic table to classify the elements potassium (K) ,Bromine (Br) and argon, (Ar) according to know likely their

atoms are to do the following
A-Lose electrons to form positive ions
B-Gain electrons to form negative ions
C-Neither gain or lose electrons
Physics
1 answer:
kodGreya [7K]3 years ago
8 0
Potassium-A
Bromine-B
Argon-C
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#86 can an object be moving when its acceleration is zero? can an object be accelerating when its speed is zero?
liraira [26]
Yes, an object<span> that was set in motion in the past by some force, but that is no longer being acted on by a net force, is </span>moving<span> but with </span>zero acceleration<span>, i.e. it is </span>moving<span> at constant velocity.</span>
7 0
3 years ago
Initially, a particle is moving at 5.25 m/s at an angle of 35.5° above the horizontal. Three seconds later, its velocity is 6.0
ivolga24 [154]

Answer:

 a =( -0.32 i ^ - 2,697 j ^)  m/s²

Explanation:

This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.

Break down the speeds in two moments

initial

  v₀ₓ = v₀ cos θ

  v₀ₓ = 5.25 cos 35.5

v₀ₓ = 4.27 m / s

   v_{oy} = v₀ sin θ

 v_{oy}= 5.25 sin35.5

v_{oy} = 3.05 m / s

Final

vₓ = 6.03 cos (-56.7)

vₓ = 3.31 m / s

v_{y} = v₀ sin θ

v_{y} = 6.03 sin (-56.7)

v_{y} = -5.04 m / s

Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order

    a = (v_{f} - v₀) /t

    aₓ = (3.31 -4.27)/3

    aₓ = -0.32 m/s²

    a_{y} = (-5.04-3.05)/3

   a_{y} =  -2.697 m/s²

6 0
3 years ago
What is the average velocity of atoms in 1.00 mol of neon (a monatomic gas) at 465 K? For m, use 0.0202 kg.
TiliK225 [7]

Answer: 757m/s

Explanation:

Given the following :

Mole of neon gas = 1.00 mol

Temperature = 465k

Mass = 0.0202kg

Using the ideal gas equation. For calculating the average kinetic energy molecule :

0.5(mv^2) = 3/2 nRt

Where ;

M = mass, V = volume. R = gas constant(8.31 jK-1 mol-1, t = temperature in Kelvin, n = number of moles

Plugging our values

0.5(0.0202 × v^2) = 3/2 (1 × 8.31 × 465)

0.0101 v^2 = 5796.225

v^2 = 5796.225 / 0.0101

v^2 = 573883.66

v = √573883.66

v = 757.55109m/s

v = 757m/s

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trapecia [35]
The answer is 44.323 cm
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