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1 answer:
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Answer:
9.96 m/s
Explanation:
mass of car, m = 487 kg
radius of track, R = 53.3 m
coefficient of static friction, μ = 0.19
acceleration due to gravity, g = 9.8 m/s^2
let v be the maximum speed so that the car can go without flying off the track.
The formula for the maximum speed is given by
vmax = 9.96 m/s
Answer:
F = 789 Newton
Explanation:
Given that,
Speed of the car, v = 10 m/s
Radius of circular path, r = 30 m
Mass of the passenger, m = 60 kg
To find :
The normal force exerted by the seat of the car when the it is at the bottom of the depression.
Solution,
Normal force acting on the car at the bottom of the depression is the sum of centripetal force and its weight.
N = 788.6 Newton
N = 789 Newton
So, the normal force exerted by the seat of the car is 789 Newton.
Answer:
j
Explanation:
x = 4 t^2 - 2 t - 4.5
Position at t = 3 s
x = 4 (3)^2 - 2 (3) - 4.5 = 25.5 m
Velocity at t = 3 s
v = dx / dt = 8 t - 2
v ( t = 3 s) = 8 x 3 - 2 = 22 m/s
Acceleration at t = 3 s
a = dv / dt = 8
a ( t = 3 s ) = 8 m/s^2
When is the velocity = 0
v = 0
8 t - 2 = 0
t = 0.25 second
When is the position = 0
x = 0
4 t^2 - 2 t - 4.5 = 0
t = 1.4 second