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PilotLPTM [1.2K]
3 years ago
6

Real life example for each gas law

Physics
1 answer:
Dvinal [7]3 years ago
3 0
Applications of Gas Law in Real Life. A torch used to heat up the and rise the air temperature inside the balloon. This cause the air volume inside the balloon to increased and becoming less dense than the surrounding air. ... The air in the ears will change its volume then causes yours ears to pop due to the strain.
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The answer is wave B has a lower frequency.  Suppose these waves represent the sound of a siren on a passing ambulance. Wave B (lower frequency) represents the sound of the siren after it has passed you.  The ambulance moving towards the observer produces a higher than normal frequency, and the ambulance moving away produces a lower than normal frequency.
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An archer fires and arrow while standing atop a 5.15 m tall wall. The arrow is fired at an angle of 55 degrees and has a launch
il63 [147K]

Answer:

Explanation:

Given

height of wall=5.15 m

angle of launch(\theta )=55^{\circ}

Launch velocity(u)=52.4 m/s

Time of flight will be sum of time of flight of projectile+time to cover 5.15 m

Time of flight of arrow=\frac{2usin\theta }{g}

t=\frac{2\times 52.4\times sin55}{9.81}=8.76 s

Now time require to cover 5.15 m

Here at the time of zero vertical displacement of arrow i.e. when arrow is at the same height as of building then its vertical velocity will change its sign compared to initial vertical velocity.

v_y=-52.4sin55 at zero vertical displacement

Thus time required will be t_2

5.15=52.4sin55\times t+\frac{gt^2}{2}

4.9t^2+42.92t-5.15=0

t=0.118 s i.e. t_2=0.118 s

total time =t_1+t_2=8.76+0.118=8.878 s

(b)Horizontal distance=Range of arrow(R_1) + horizontal distance in 0.118 s(R_2)

R_1=\frac{u^2sin2\theta }{g}=\frac{52.4^2\times sin110}{9.8}=263.28 m

R_2=52.4cos55 \times 0.118=3.546 m

R=R_1+R_2=263.28+3.546=266.82 m

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ρ = 4.05 g  / 12 mL

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<h3>ρ ≅ 0.338 g/mL</h3>

OPTION A

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