Answer: 40.84 m
Explanation:
Given
Radius of the disk, r = 2m
Velocity of the disk, v = 7 rad/s
Acceleration of the disk, α = 0.3 rad/s²
Here, we use the formula for kinematics of rotational motion to solve
2α(θ - θ•) = ω² - ω•²
Where,
ω• = 0
ω = v/r = 7/2
ω = 3.5 rad/s
2 * 0.3(θ - θ•) = 3.5² - 0
0.6(θ - θ•) = 12.25
(θ - θ•) = 12.25 / 0.6
(θ - θ•) = 20.42 rad
Since we have both the angle and it's radius, we can calculate the arc length
s = rθ = 2 * 20.42
s = 40.84 m
Thus, the needed distance is 40.84 m
Answer:
4992 J
Explanation:
Force, F = 39 N
Distance, s = 128 m
Work done = force x distance
W = 39 x 128
W = 4992 J
Thus, the work done is 4992 J.
Answer:
103063860 Pa
Explanation:
= Density of seawater = 1030 kg/m³
g = Acceleration due to gravity = 9.81 m/s²
h = Depth at which pressure is being measured = 10.2 km
The gauge pressure is given by
Therefore, the gauge pressure at a depth of 10.2 km is 103063860 Pa
The answer is 3 m. This is the area under the graph from t=2 to t=3, using the trapezium rule. 1/2 (2+4) * 1