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alexandr402 [8]
3 years ago
7

Heat is added to an element causing the atoms to bounce back and forth in the molecule. This is an example of which type of moti

on? Question 2 options: Rotational motion Vibrational motion Translational motion Inertia asap
Physics
2 answers:
lisabon 2012 [21]3 years ago
8 0

Answer:

Vibrational motion

Explanation:

jok3333 [9.3K]3 years ago
6 0

On adding heat to a molecule, atoms start bouncing back and forth and repeat the motion that are known as the vibrations.

Answer: Option B

<u>Explanation: </u>

We often see telephone wires getting expanded and hung up lower in the hot summer days while they are much more contracted during the winter times. This is because of the molecular expansion as when the atoms of the molecules gain sufficient energy, they start vibrating back and forth from their median position and start expanding their vibration spaces.

Due to this, the molecular structure of the object also expand and hence the object. On the other hand, when heat is extracted from the object to the outer atmosphere, atoms contract in their spaces of vibrations and hence, the molecular structure also contract. This means, the object contract or shrink from their original size and structure when heat is extracted from them.

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2 years ago
When Sr and F combine to form a compound, which atom(s) will become an anion?\
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Explanation:

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7 0
3 years ago
Complete all four parts. 15 points. Will give brainliest! Show work!
Vlada [557]

Answer:

A. 5.08 secs.

B. 10.16 secs.

C. 126.50 m.

D. 373.36 m

Explanation:

Data obtained from the question include the following:

Initial velocity (u) = 65 m/s

Angle of projection θ = 50°

A. Determination of the time taken to reach the peak.

Initial velocity (u) = 65 m/s

Angle of projection θ = 50°

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =.?

t = u•Sine θ/g

t = (65 × Sine 50) /9.8

t = 5.08 secs.

B. Determination of the total time spent by the ball in air.

Time (t) taken to reach the peak = 5.08 secs.

Total time (T) spent by the ball in air =?

T = 2t

T = 2 × 5.08

T = 10.16 secs

Therefore, the total time spent by the ball in air is 10.16 secs.

C. Determination of the maximum height.

Initial velocity (u) = 65 m/s

Angle of projection θ = 50°

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (H) =..?

H = u²•Sine² θ / 2g

H = 65² × (Sine 50)² / 2 × 9.8

H = 4225 × (Sine 50)² /19.6

H = 126.50 m

Therefore, the maximum height reached by the ball is 126.50 m.

D. Determination of the horizontal distance travelled by the ball.

Initial velocity (u) = 65 m/s

Angle of projection θ = 50°

Acceleration due to gravity (g) = 9.8 m/s²

Horizontal distance (R) =..?

R = u²•Sine 2θ / g

R = 65² × Sine (2×30) / 9.8

R = (4225 × Sine 60) / 9.8

R = 373.36 m

Therefore, the horizontal distance travelled by the ball is 373.36 m

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