the answer is 0.44 rad/sec.
This is true because the equation for angular velocity is w = (theta)/(time). Since we know theta of a circle is 2pi, and the time is 14.3 seconds, we can just plug the numbers in.
w = 2pi / 14.3 = 0.44 rad/sec.
This is true because the equation for angular velocity is w = (theta)/(time). Since we know theta of a circle is 2pi, and the time is 14.3 seconds, we can just plug the numbers in.
w = 2pi / 14.3 = 0.44 rad/sec.
a = (v^2) / r
Where:
a = acceleration
v = velocity
r = radius
We can find the velocity of the car by dividing the circumference of the circular track by the time it takes to complete one lap:
v = (2Ï€r) / t
Let's calculate the values step by step.
Step 1: Calculate the velocity of the car.
The circumference of the circular track can be calculated using the formula 2Ï€r:
C = 2Ï€r
C = 2Ï€(45.0 m)
C ≈ 282.743 m
Now, we can calculate the velocity using the formula v = C / t:
v = 282.743 m / 14.3 s
v ≈ 19.8 m/s
Step 2: Calculate the acceleration of the car.
Using the formula a = (v^2) / r:
a = (19.8 m/s)^2 / 45.0 m
a ≈ 8.73 m/s^2
Therefore, the acceleration of the car is approximately 8.73 m/s².
Step 3: Calculate the force exerted by the track on the tires.
The force exerted by the track on the tires is equal to the mass of the car multiplied by the acceleration:
F = m * a
F = 590 kg * 8.73 m/s^2
F ≈ 5140.7 N
Therefore, the track must exert a force of approximately 5140.7 N on the tires to produce this acceleration.
acceleration = (2 * π * radius) / time^2
(A) Plugging in the values, we get:
acceleration = (2 * π * 45.0 m) / (14.3 s)^2
Simplifying this calculation gives us:
acceleration ≈ 1.98 m/s²
Therefore, the acceleration of the car is approximately 1.98 m/s².
(B) To find the force that the track must exert on the tires, we can use Newton's second law of motion:
force = mass * acceleration
Plugging in the given mass and calculated acceleration:
force = 590 kg * 1.98 m/s²
The calculation gives us:
force ≈ 1168.2 N
Therefore, the track must exert a force of approximately 1168.2 Newtons on the tires to produce this acceleration.