The school that was supported was the Self psychology or the Ego psychology which is based on the model of the mind called id-ego-superego model
Answer:
Assume that the ball undergoes motion along a straight line. ... F = m A Force = (mass) x (acceleration) The question tells you the mass and the acceleration. All YOU have to do is take the numbers and pluggum into Newton's 2nd law. F = m A = (0.75 kg) (25 m/s²) = (0.75 x 25) kg-m/s² = 18.75 Newtons .
Explanation:
i looked it up ok
The Voyager and Pioneer flybys of the 1970s and 1980s provided rough sketches of Saturn’s moons. But during its many years in Saturn orbit, Cassini discovered previously unknown moons, solved mysteries about known ones, studied their interactions with the rings and revealed how sharply different the moons are from one another.
The answer is not 'A'.
Using the numbers given in the question, it's 'B'.
It WOULD be 'A' if the number in B were 71 or less.
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<span>This is not as simple as it looks.
What quantity are we going to compare between the two cases ?
Yes, I know ... the "amount of work". But how to find that from the
numbers given in the question ?
Is it the same as the change in speed ?
Well ? Is it ?
NO. IT's NOT.
In order to reduce the car's speed, the brakes have to absorb
the KINETIC ENERGY, and THAT changes in proportion to
the SQUARE of the speed. ( KE = 1/2 m V² )
Case 'A' :
The car initially has (1/2 m) (100²)
= (1/2m) x 10,000 units of KE.
It slows down to (1/2 m) x (70²)
= (1/2m) x 4,900 units of KE.
The brakes have absorbed (10,000 - 4,900) = 5,100 units of KE.
Case 'B' :
The car initially has (1/2 m) (79²)
= (1/2m) x 6,241 units of KE.
It slows down to a stop . . . NO kinetic energy.
The brakes have absorbed all 6,241 units of KE.
Just as we suspected when we first read the problem,
the brakes do more work in Case-B, bringing the car
to a stop from 79, than they do when slowing the car
from 100 to 70 .
But when we first read the problem and formed that
snap impression, we did it for the wrong reason.
Here, I'll demonstrate:
Change Case-B. Make it "from 71 km/h to a stop".
Here's the new change in kinetic energy for Case-B:
The car initially has (1/2 m) (71²)
= (1/2m) x 5,041 units of KE.
It slows down to a stop . . . NO kinetic energy.
The brakes have absorbed all 5,041 units of KE.
-- To slow from 100 to 70, the brakes absorbed 5,100 units of KE.
-- Then, to slow the whole rest of the way from 71 to a stop,
the brakes absorbed only 5,041 units of KE.
-- The brakes did more work to slow the car the first 30 km/hr
than to slow it to a complete stop from 71 km/hr or less.
That's why you can't just say that the bigger change in speed
requires the greater amount of work.
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It works exactly the same in the opposite direction, too.
It takes less energy from the engine to accelerate the car
from rest to 70 km/hr than it takes to accelerate it the
next 30, to 100 km/hr !</span>
(The exact break-even speed for this problem is 50√2 km/h,
or 70.711... km/hr rounded. )
The normal acceleration in the United States is zero to sixty time. The average rate on ordinary cars was between 3 and 4 m/s2
