An element which is highly conductive, highly reactive, soft, and lustrous is most likely an alkali metal.
Alkali metals are in group 1 of the Periodic table which means that they have only a single valence electron.
This causes them to be soft and highly reactive because:
- The single valance electron leads to weak bonds amongst the element's atoms which makes them soft
- The elements want to lose the single valance electron so as to become stable so they will react with other elements to give away the electron.
Examples of alkali electrons include:
- Lithium
- Sodium
- Potassium etc
In conclusion therefore, alkali metals are highly reactive and soft and so the element described above is most likely an alkali metal.
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Answer: see the graph attached (straight line, passing through the origin and positive slope).
Justification:1)
Kinetic energy and temperature are in direct proportion. That means:
i) Being kinetic energy y and temperature x:
y α xii) That implies:
y = kx,where k is the constant of proportionality.
iii) The graph is a
line that passes through the origin and has positive slope k (k = y / x).2) The proportional relationship between kinetic energy (KE) and temperature (T) is shown by the
Boltzman law, which states:
Average KE = [3 / 2] KT, where K is Boltzman's constant, whose graph is of the form shown in the figure attached.
Answer:
v = 29.4m/s
Explanation:
Since the ball is dropped at rest,
u = 0m/s
a = 9.81m/s²
Using
v = u + at
After 3 seconds,
v = 0 + (9.81)(3)
v = 29.4m/s
Explanation:
The magnitude of a vector v can be found using Pythagorean's theorem.
||v|| = √(vₓ² + vᵧ²)
||v|| = √((-309)² + (187)²)
||v|| ≈ 361
You can find the angle of a vector using trigonometry.
tan θ = vᵧ / vₓ
tan θ = 187 / -309
θ ≈ 149° or θ ≈ 329°
vₓ is negative and vᵧ is positive, so θ must be in the second quadrant. Therefore, θ ≈ 149°.
My guess is either they come to a stop when they come in contact, or one ball is going to go the opposite direction.
