Answer:
139.514 metres
Explanation:
Initial velocity of the truck = 6.6 m/s
Acceleration of the truck = 2.8 m/s^2
Time interval = 7.9 s
Therefore we use the formula,
s = ut + 1/2 at^2
*where s(the distance travelled)...u(the initial velocity)...t(the time period)
; s = 6.6(7.9) + 1/2 (2.8)(7.9)^2
; s = 52.14 + 87.374
The distance moved by the truck = 139.514m
To find the force we use the formula,
F = ma , where m is mass and a acceleration
Using the formula,
F = ma
F = 0.42 x 14.8
F = 6.216 N / 6.22 N
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Answer:
0.42 m/s²
Explanation:
r = radius of the flywheel = 0.300 m
w₀ = initial angular speed = 0 rad/s
w = final angular speed = ?
θ = angular displacement = 60 deg = 1.05 rad
α = angular acceleration = 0.6 rad/s²
Using the equation
w² = w₀² + 2 α θ
w² = 0² + 2 (0.6) (1.05)
w = 1.12 rad/s
Tangential acceleration is given as
= r α = (0.300) (0.6) = 0.18 m/s²
Radial acceleration is given as
= r w² = (0.300) (1.12)² = 0.38 m/s²
Magnitude of resultant acceleration is given as


= 0.42 m/s²
38 = 300
60 = ?
300 × 60 ÷ 38
1800 ÷ 38 = 47.368
The answer is 47.4 m per hour