The complementary color of a red-violet focal point is <span>yellow-green.
The complementary colors are </span>colors which appear opposite each other on the color wheel. These colors complete each other on the color spectrum. <span>Complementary colors have the ability to increase intensities when used next to each other.</span>
Answer:
0.12m/s
Explanation:
v=λf
Given that, λ = 12cm = 0.12m
T = 1second
(A period T is the time required for one complete cycle of vibration to pass a given point)
frequency 'f' is unknown but we can get frequency from f = 1/T = 1/1 = 1Hz
therefore, v= 0.12 × 1 = 0.12m/s
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. )