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Ghella [55]
3 years ago
13

__________ brake systems release the brakes momentarily when wheel speed sensors indicate a locked wheel during braking and trac

tion. Control systems apply the brakes momentarily to one of the drive wheels whenever the wheel speed sensors indicate a wheel is going faster than the others during acceleration.
Physics
2 answers:
nasty-shy [4]3 years ago
6 0

Answer:

Anti- Lock

Explanation:

Anti- lock brake systems(ABS) is a safety system used in hindering the wheels from locking up when the brake is applied.

Braking release the brakes momentarily when wheel speed sensors indicate a locked wheel during braking and traction.

Control systems apply the brakes momentarily to one of the drive wheels whenever the wheel speed sensors indicate a wheel is going faster than the others during acceleration.

chubhunter [2.5K]3 years ago
5 0

"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.

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A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a
Rasek [7]

Answer:

W=173.48J

Explanation:

information we know:

Total force: F=45N

Weight: w=100N

distance: 4m

vertical component of the force: F_{y}=12N

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In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: F_{x}=Fcos\theta

vertical component: F_{y}=Fsen\theta

but from the given information we know that F_{y}=12N

so, equation these two F_{y}=Fsen\theta and F_{y}=12N

Fsen\theta =12N

and we know the force F=45N, thus:

45sen\theta=12

now we clear for \theta

sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

F_{x}=Fcos\theta

F_{x}=45cos(15.466)\\F_{x}=43.37N

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J

the work done is W=173.48J

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