The final temperature of the gas given the data from the question is 1527 °C
<h3>Data obtained from the question</h3>
- Initial volume (V₁) = V
- Initial temperature (T₁) = 27 °C = 27 + 273 = 300 K
- Final volume (V₂) = 6V
- Final temperature (T₂) =?
<h3>How to determine the new temperature </h3>
The final temperature of the gas can be obtained by using the Charles' law equation as illustrated below:
V₁ / T₁ = V₂ / T₂
V / 300 = 6V / T₂
Cross multiply
V × T₂ = 300 × 6V
Divide both side by V
T₂ = (300 × 6V) / V
T₂ = 1800 K
Subtract 273 from 1800 K to express in degree celsius
T₂ = 1800 – 273
T₂ = 1527 °C
Learn more about gas laws:
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Answer: Longitudinal wave
Explanation:
Longitudinal wave are the oscillations that are parallel to the direction of energy transfer that means the vibrations are in line with the direction where the energy is travelling.
A key feature of sound wave is that they cause sound particles to vibrate. The region where the particles are close together are called compressions and regions where particles are further apart they are called rarefactions.
The other options explanation:
-Transverse waves are where the oscillations are perpendicular to the energy of transfer.
-A standing wave is where the waves are travelling back and forth where there are some fixed points in the system whilst other vibrate with highest amplitude
-Surface waves have both the characteristics of longitudinal and transverse waves
Explanation:
First convert the speed into m/s and time into seconds
Answer:
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Explanation:
<u>Net Force And Acceleration
</u>
The Newton's second law relates the net force applied on an object of mass m and the acceleration it aquires by

The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N
The components of the first force are


The components of the second force are


The net force is


The magnitude of the net force is


The acceleration has a magnitude of



The direction of the acceleration is the same as the net force:

