Answer:
a) t = 0.75 s, b) t = 2.25 s
Explanation:
The speed of sound is constant in a material medium
v = 340 m / s
we can use the relations of uniform motion to find the time
v = x / t
t = x / v
In the exercise, the observer's distance to the wall is indicated d = 510 m, it also indicates that the shot is fired at the midpoint
x = d / 2
a) direct sound the distance from the observer to the screen is
x = 510/2 = 255 m
t = 255/340
t = 0.75 s
b) echo sound.
In this case the sound reaches the wall bounces, the distance is
x = d / 2 + d
x = 3/2 d
x = 3/2 510
x = 765 m
the time is
t = x / v
t = 765/340
t = 2.25 s
Answer:
The initial volume of the system is 2.7 L
Explanation:
The change in internal energy is given by;
ΔU = Q - W
where;
Q is the heat absorb from the surrounding
W is the work done by the system
Given;
ΔU = -107.6 J
Q = 59 J
W = ?
W = Q- ΔU
W = 59J - (-107.6)
W = 166.6 J
Work done is given by;
W = PΔV
W = P(V₂ - V₁)
101.325 Joules = 1 L.atm
166.6 Joules = ?
= 1.644 L.atm
1.644 L.atm = 0.543 atm(57.7L - V₁)
57.7L - V₁ = (1.644 L.atm) / (0.543 atm)
57.7L - V₁ = 3.0 L
V₁ = 57.7L - 3.0 L
V₁ = 2.7 L
Therefore, the initial volume of the system is 2.7 L
Answer:
its C! I just finished the test on edg :)
