Answer:
0.54m
Explanation:
Step one:
given data
length of seesaw= 3m
mass of man m1= 85kg
weight = mg
W1= 85*10= 850N
mass of daughter m2= 35kg
W2= 35*10= 350N
distance from the center= (1.5-0.2)= 1.3m
Step two:
we know that the sum of clockwise moment equals the anticlockwise moment
let the distance the must sit to balance the system be x
taking moment about the center of the system
350*1.3=850*x
455=850x
divide both sides by 850
x=455/850
x=0.54
Hence the man must sit 0.54m from the right to balance the system
Verrrrry interesting !
If the moon were replaced by something with a vastly greater mass
but at the same distance, then ...
-- The period of its revolution around the Earth would be much shorter.
That is, it would orbit the Earth in much less than 27.3 days. We might
see it go through a complete set of phases in 2 weeks, or even 1 week.
-- The ocean tides would be much greater. Low tides would be
much lower, and high tides would be much higher.
-- Sadly, the land tides, and the forces on the Earth's internal structure,
would also be much greater. That means great increases in earthquake
and volcanic activity.
-- The Earth and moon both revolve around their common center of
mass. Under the current arrangement ... with the Earth having 80 times
the mass of the Moon ... that point is inside the Earth, and it looks a lot
like the Moon is orbiting a stationary Earth.
When the new body arrives to replace the lightweight Moon, that point
will be a lot closer to the new companion ... maybe even inside it.
Then, it will look a lot like the monster is the stationary one, and the
Earth is orbiting it.
I actually don't believe that we would SEE that change, or feel it.
Kinesthetic learners learn best "hands-on" to answer your question the answer would be B.
Reflection,polarization,interference patterns
Answer:
Car A is 1.8-1.2= 0.6 times faster than car B
So the distance between them is changing 0.6 times faster.
Explanation:
Car A is traveling north at 90km/hr and 0.3 km away from intersection point.
Car b is traveling west at 80km/hr and 0.4 km away from the intersection point.
The time it will take to get to intersection point by those two cars is how fast it is changing.
For car A
Time = distance/speed
Time = 0.3/90
Time = 3.333*10^-3
Time = 3.333*10^-3 hrs
Converting to seconds
(3.333*10^-3)*360= 1.199998
Approximately 1.2 seconds
For car B
Time = distance/time
Time= 0.4/80
Time = 5*10^-3
Time= 5*10^-3 hrs
Converting to seconds
(5*10^-3)*360= 1.8 seconds
Comparing the two times,
Car A is 1.8-1.2= 0.6 times faster than car B