<span>For this example, the value presented would be considered a statistic. The value is a statistic as it represents a numerical measurement of a sample. If it were a parameter, it would need to represent a numerical measurement of a population.</span>
Answer: The final temperature is 470K
Explanation: Using the relation;
Q= ΔU +W
Given, n = 2mol
Initial temperature T1= 345K
Heat =Q= 2250J
Workdone=W=-870J(work is done on gas)
T2 =Final temperature =?
ΔU =3/2nR(T2-T1)
ΔU=3/2 × 2 ×8.314 (T2 - 345)
ΔU=24.942(T2-345)
Therefore Q = 24.942(T2-345)+ (-870)
2250=24.942(T2-345)+ (-870)
125.09=(T2-345)
T2 =470K
Therfore the final temperature is 470K
Answer:
nope dont agree with that i think it would b a lot harder to do on a mass scale like that
Explanation:
Only way to do that is if aliens with far superior technology wise came to earth and did it
Answer:
269 m
45 m/s
-58.6 m/s
Explanation:
Part 1
First, find the time it takes for the package to land. Take the upward direction to be positive.
Given (in the y direction):
Δy = -175 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(-175 m) = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 5.98 s
Next, find the horizontal distance traveled in that time:
Given (in the x direction):
v₀ = 45 m/s
a = 0 m/s²
t = 5.98 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²
Δx = 269 m
Part 2
Given (in the x direction):
v₀ = 45 m/s
a = 0 m/s²
t = 5.98 s
Find: v
v = at + v₀
v = (0 m/s²) (5.98 s) + (45 m/s
v = 45 m/s
Part 3
Given (in the y direction):
Δy = -175 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (-9.8 m/s²) (-175 m)
v = -58.6 m/s
