4.) In coming to a stop, a car leaves skid marks 80m long on the highway. Assuming a DEceleration of 7.00 m/s^2 (m/s/s), estimate the speed of the car just before breaking.
6.) A car traveling @ 90km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80m. What was the average acceleration of the driver during the collision? Express answer in terms of ("g's") where 1.00 g = 9.80 m/s^2 OR (m/s/s)
A car is traveling at a speed of 80km/h on an icy highway. The driver swerves while applying the brakes to avoid a collision. if the Yaw mark that the car leaves on the icy asphalt has a radius of 270 m, determine the drag factor of the icy highway. Round to the nearest hundredth.
To solve for the drag factor of the icy highway, we can use the equation:
drag factor = (velocity^2) / (radius * acceleration)
First, let's convert the velocity from km/h to m/s:
80 km/h = 80 * (1000 m / 3600 s) = 22.22 m/s
Next, let's plug in the values into the equation:
drag factor = (22.22^2) / (270 * acceleration)
To determine the acceleration, we need to make use of the centripetal acceleration formula:
acceleration = (velocity^2) / radius
acceleration = (22.22^2) / 270
Now we can substitute the acceleration value back into the drag factor equation:
drag factor = (22.22^2) / (270 * [(22.22^2) / 270])
Simplifying:
drag factor = 1
Therefore, the drag factor of the icy highway is 1.00.
s = sqrt127fr
Where,
s = the speed (km/h) the vehicle was travelling when the brakes were applied
f = the drag factor of the surface
r = radius (m) of the curved Yaw mark
A car is traveling at a speed of 80km/h on an icy highway. The driver swerves while applying the brakes to avoid a collision. if the Yaw mark that the car leaves on the icy asphalt has a radius of 270 m, determine the drag factor of the icy highway. Round to the nearest hundredth.
To find the drag factor of the icy highway, we can use the equation:
s = √(127f * r)
Given that the speed (s) is 80 km/h and the radius (r) of the yaw mark is 270 m, we can rearrange the equation to solve for the drag factor (f):
f = (s^2) / (127 * r)
First, let's convert the speed from km/h to m/s:
s = 80 * (1000 m / 3600 s)
s = 22.22 m/s
Now we can plug in the values into the equation:
f = (22.22^2) / (127 * 270)
f ≈ 0.045
Therefore, the drag factor of the icy highway is approximately 0.045.
sqrt8
The square root of 8 is approximately 2.828.
entire radical next to its equivalent mixed radical is 2sqrt3
To rewrite √8 as a mixed radical, we can break it down into its simplest form by factoring out perfect square factors from the radicand.
√8 = √(4 * 2)
Since 4 is a perfect square, we can take its square root out of the radical:
√(4 * 2) = 2√2
Therefore, the expression 2√3 is not equal to √8.
9 square root of 14 and square root of 14 are like terms. select one true or false
False.
9√14 and √14 are not like terms because they have different coefficients. Like terms have the same variable(s) and exponent(s), but the coefficients can be different. In this case, the coefficient of 9 in 9√14 makes it different from √14.