Answer:
the value of the final pressure is 0.168 atm
Explanation:
Given the data in the question;
Let p₁ be initial pressure, v₁ be initial volume.
After expansion, p₂ is final pressure and v₂ is final volume.
So using the following equations;
p₁v₁ = nRT
p₂v₂ = nRT
hence, p₁v₁ = p₂v₂
we find p₂
p₂ = p₁v₁ / v₂
given that; initial volume v₁ = 0.175 m³, Initial pressure p₁ = 0.350 atm,
final volume v₂ = 0.365 m³
we substitute
p₂ = ( 0.350 atm × 0.175 m³ ) / 0.365 m³
p₂ = 0.06125 atm-m³ / 0.365 m³
p₂ = 0.168 atm
Therefore, the value of the final pressure is 0.168 atm
Answer:
A friend snorkeling just below the surface of the water along the same shore will detect the sound first.
Explanation:
- The speed of sound in water medium is faster than that through the air.
- Sound propagates through the medium by transferring through the molecules on it. Water has more closely packed molecules due to which the speed is faster.
- In fact, the sound's speed in water is almost four times faster than that in the air.
- So the guy in the water surface gets to hear sound faster than the one in sore.
Answer:
W = 0.060 J
v_2 = 0.18 m/s
Explanation:
solution:
for the spring:
W = 1/2*k*x_1^2 - 1/2*k*x_2^2
x_1 = -0.025 m and x_2 = 0
W = 1/2*k*x_1^2 = 1/2*(250 N/m)(-0.028m)^2
W = 0.060 J
the work-energy theorem,
W_tot = K_2 - K_1 = ΔK
with K = 1/2*m*v^2
v_2 = √2*W/m
v_2 = 0.18 m/s
Risks are only in a persons
