"Anti-Lock" brake systems release the brakes momentarily when wheel speed sensors indicate a locked wheel during braking and traction.
<u>Explanation:</u>
The safety anti-skid braking system is known as "anti-lock braking system" having huge application on land vehicles like one, two and multiple wheeler vehicles and aircraft. During braking, it avoids wheels to get locked by building tractive contacts to the road's surface.
This seems to be an automated system work on the principles of techniques - threshold and cadence braking. The wheel velocity sensors are utilized by ABS to find whether one or more than one wheels chose to get lock while braking.
<span>equal-arm balance = 12 kg
spring scale = 2 kg
The equal arm balance measures the mass of an object by using a counter mass on the opposite plate of the balance. The force of gravity affects both the mass being tested and the mass standards being compared against equally. So if the local gravitational field changes, those changes affect both the tested mass and the standard masses equally.
Contrast that to the spring scale. In that scale, the spring provides a calibrated level of force irrespective of the local gravitational field. So if the local gravity is higher, the force indicated also is higher. And if the gravitational field is lower, the indicated force is also lower. The strength of the spring DOES NOT CHANGE with changes in the local gravitational field.</span>
Answer:
first order date and most recent order date
Explanation:
it was switched. column 5 should be most recent order date because it's 2020 while column 6 should be first order date because it was in 2019
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Answer:
0.32 m.
Explanation:
To solve this problem, we must recognise that:
1. At the maximum height, the velocity of the ball is zero.
2. When the velocity of the ball is 2.5 m/s above the ground, it is assumed that the potential energy and kinetic energy of the ball are the same.
With the above information in mind, we shall determine the height of the ball when it has a speed of 2.5 m/s. This can be obtained as follow:
Mass (m) = constant
Acceleration due to gravity (g) = 9.8 m/s²
Velocity (v) = 2.5 m/s
Height (h) =?
PE = KE
Recall:
PE = mgh
KE = ½mv²
Thus,
PE = KE
mgh = ½mv²
Cancel m from both side
gh = ½v²
9.8 × h = ½ × 2.5²
9.8 × h = ½ × 6.25
9.8 × h = 3.125
Divide both side by 9.8
h = 3.125 / 9.8
h = 0.32 m
Thus, the height of the ball when it has a speed of 2.5 m/s is 0.32 m.
