An ideal gas expands at a constant total pressure of 3.0 atm from 400 mL to 600 mL. heat then flows out of the gas at a constant
1 answer:
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Answer:
Vi = 32 [m/s]
Explanation:
In order to solve this problem we must use the following the two following kinematics equations.
The negative sign of the second term of the equation means that the velocity decreases, as indicated in the problem.
where:
Vf = final velocity = 8[m/s]
Vi = initial velocity [m/s]
a = acceleration = [m/s^2]
t = time = 5 [s]
Now replacing:
8 = Vi - 5*a
Vi = (8 + 5*a)
As we can see we have two unknowns the initial velocity and the acceleration, so we must use a second kinematics equation.
where:
d = distance = 100[m]
(8^2) = (8 + 5*a)^2 - (2*a*100)
64 = (64 + 80*a + 25*a^2) - 200*a
0 = 80*a - 200*a + 25*a^2
0 = - 120*a + 25*a^2
0 = 25*a(a - 4.8)
therefore:
a = 0 or a = 4.8 [m/s^2]
We choose the value of 4.8 as the acceleration value, since the zero value would not apply.
Returning to the first equation:
8 = Vi - (4.8*5)
Vi = 32 [m/s]
Answer:
T=26.03 N
Explanation:
Given that
Distance between poles = 12 m
Mass of block m= 4 kg
Sag distance = 5 m
Lets take tension in the clothesline is T.
The component of tension in vertical direction will be T cosθ.
By force balancing
2 T cosθ = 40
here
θ=39.80°
2 T cos39.8 = 40
T=26.03 N
