let the distance of pillar is "r" from one end of the slab
So here net torque must be balance with respect to pillar to be in balanced state
So here we will have

here we know that
mg = 19600 N
Mg = 400,000 N
L = 20 m
from above equation we have



so pillar is at distance 10.098 m from one end of the slab
Answer:
C). Take your foot off the gas pedal. Then brake lightly until you are moving at low speed.
Explanation:
While driving on roads of rural areas, if our right wheel moves off the pavement, we should always hold the steering wheel firmly and then take our foot off the gas pedal, then apply brake lightly until we are moving at a low speed.
When our wheels drift off the pavement area, we should not panic and yank. And instead of turning the wheel back in the left direction towards the road, it is always safer to take off our foot from the gas pedal and then apply brakes slowly. When our vehicle slows down check the incoming traffic behind us and then we should slowly move back on to the pavement.
Answer:
The inter-molecular forces holding non-polar compounds together is low compare to that of polar compounds. Therefore, it will take less energy to break the bond for non-polar compounds and vice versa. That is why polar compounds have higher melting points than non-polar compounds.
Explanation:
The final speed of the car at the given conditions is 30.1 m/s.
The given parameters:
- <em>Mass of the car, m = 1700 kg</em>
- <em>Velocity of the car, v = 21 m/s</em>
- <em>Time of motion, t = 10 s</em>
- <em>Additional energy provided by the engine, E₁ = 22,000 J</em>
- <em>Energy used in overcoming friction, E₂ = 3,666.67 J</em>
The change in the energy applied to the car is calculated as;

The final speed of the car is calculated as follows;

Thus, the final speed of the car at the given conditions is 30.1 m/s.
Learn more about change in kinetic energy here: brainly.com/question/6480366