Answer:
The fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C
Explanation:
Given;
initial temperature of the liquid, t₁ = 76.3 +/- 0.4⁰C
final temperature of the liquid, t₂ = 67.7 +/- 0.3⁰C
The change in temperature of the liquid is calculated as;
Δt = t₂ - t₁
Δt = (67.7 - 76.3) +/- (0.3 - 0.4)
Δt = (-8.6) +/- (-0.1)
Δt = 8.6 +/- 0.1 ⁰C
Therefore, the fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C
Answer:
The rotor's angular velocity is 82.73rad/s
Explanation:
It is a curvilinear movement of a constant radius. If there is uniform angular acceleration, then it is a circular motion with constant acceleration, whose equations are analogous to that of the translational motion.
Calculating the initial velocity of the rotor, V1 in rad/s
V1 = 610rev/minute × 6.28 × 1miute/60secs
V1 = 63.85rad/s
Using kinematic equation to calculate the final velocity of the rotor
Given:
Angular acceleration = 5.9rad/s^2
Time,t= 3.2seconds
V2 = V1 + a × t
V2 = 63.85 + (5.9)× (3.2)
V2 = 63.85 + 18.88
V2 = 82.73rads/s
We calculate the coordinates at t₁ = 9 min and t₂ = 10 min, since the 10th minute is between t₁ and t₂.
As it leaves from rest, it means that the initial speed is zero
t₁=9 min=540 s
t₂=10 min=600 s
x₁=at₁²/2=8*540²/2=4*291600=1166400 m
x₂=at₂²/2=8*600²/2=4*360000=1440000 m
Δx=x₂-x₁=1440000-1166400=273600 m represents the distance traveled by the car in the 10th minute of travel
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