Answer:
The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J
Explanation:
The given variables are
Work done = 550 J
Volume change = V₂ - V₁ = -0.5V₁
Thus the product of pressure and volume change = work done by gas, thus
P × -0.5V₁ = 500 J
Hence -PV₁ = 1000 J
also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂
Also to compress the gas by a factor of 11 we have
P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work
a)5m/s b)5
the 5 is because you add the seconds to get 8 seconds and then do the same with the distance to get 40. 40/8 = 5. speed = 5
Velocity = displacement/change in time
V = 40/8
I just realized how unorganised my math looks but I hope this is helpfull
Answer:Correct answer: 15.85 kg·m/s
Explanation:
A 30 kg gun is standing on a frictionless sur-face. The gun fires a 50 g bullet with a muzzlevelocity of 317 m/s.The positive direction is that of the bullet.Calculate the momentum of the bullet im-mediately after the gun was fired.
Answer:
a. It became deeper by a factor of 3.
Explanation:
What we have is water flowing down a river with constant width. The water slows from speed v to v3 over a shirt distance
Using the equation of continuity
A1V1 = A2V2 ----1
A1 is the area of rectangle
V1 is the velocity of water
Area of rectangle = length x width
We rewrite equation 1 as
λ1w1v1 = λ2w2v2
We have w1 = w2
λ1v1 = λ2v2
λ1*v1 = λ2*v/3
λ1 = λ2/3
So it becomes deeper by a factor of 3
Answer:
A. loses an electron, it becomes positively charged.
