Answer:
Crumple zones are designed to absorb and redistribute the force of a collision. ... Also known as a crush zone, crumple zones are areas of a vehicle that are designed to deform and crumple in a collision. This absorbs some of the energy of the impact, preventing it from being transmitted to the occupants.
Answer:
<em>Magnetic</em><em> </em><em>compass</em><em> </em><em>helps</em><em> </em><em>to </em><em>identify</em><em> </em><em>direction</em><em> </em><em>in </em><em>this </em><em>way </em><em>,</em><em> </em><em>this </em><em>compass</em><em> </em><em>work </em><em>because</em><em> </em><em>of </em><em>earth</em><em> </em><em>magnetic</em><em> field</em><em> </em><em>and </em><em>show</em><em> </em><em>us </em><em>direction</em><em> </em>
<em> </em><em> </em><em>hope</em><em> it</em><em> helps</em><em> and</em><em> your</em><em> day</em><em> will</em><em> be</em><em> full</em><em> of</em><em> happiness</em><em>. </em>^_^
I think the key here is to be exquisitely careful at all times, and
any time we make any move, keep our units with it.
We're given two angular speeds, and we need to solve for a time.
Outer (slower) planet:
Angular speed = ω rad/sec
Time per unit angle = (1/ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/ω sec/rad) · (2π rad) = 2π/ω seconds .
Inner (faster) planet:
Angular speed = 2ω rad/sec
Time per unit angle = (1/2ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/2ω sec/rad) · (2π rad) = 2π/2ω sec = π/ω seconds.
So far so good. We have the outer planet taking 2π/ω seconds for one
complete revolution, and the inner planet doing it in only π/ω seconds ...
half the time for double the angular speed. Perfect !
At this point, I know what I'm thinking, but it's hard to explain.
I'm pretty sure that the planets are in line on the same side whenever the
total elapsed time is something like a common multiple of their periods.
What I mean is:
They're in line, SOMEwhere on the circles, when
(a fraction of one orbit) = (the same fraction of the other orbit)
AND
the total elapsed time is a common multiple of their periods.
Wait ! Ignore all of that. I'm doing a good job of confusing myself, and
probably you too. It may be simpler than that. (I hope so.) Throw away
those last few paragraphs.
The planets are in line again as soon as the faster one has 'lapped'
the slower one ... gone around one more time.
So, however many of the longer period have passed, ONE MORE
of the shorter period have passed. We're just looking for the Least
Common Multiple of the two periods.
K (2π/ω seconds) = (K+1) (π/ω seconds)
2Kπ/ω = Kπ/ω + π/ω
Subtract Kπ/ω : Kπ/ω = π/ω
Multiply by ω/π : K = 1
(Now I have a feeling that I have just finished re-inventing the wheel.)
And there we have it:
In the time it takes the slower planet to revolve once,
the faster planet revolves twice, and catches up with it.
It will be 2π/ω seconds before the planets line up again.
When they do, they are again in the same position as shown
in the drawing.
To describe it another way . . .
When Kanye has completed its first revolution ...
Bieber has made it halfway around.
Bieber is crawling the rest of the way to the starting point while ...
Kanye is doing another complete revolution.
Kanye laps Bieber just as they both reach the starting point ...
Bieber for the first time, Kanye for the second time.
You're welcome. The generous bounty of 5 points is very gracious,
and is appreciated. The warm cloudy water and green breadcrust
are also delicious.
solution:
radius of steel ball(r)=5cm=0.05m
density of ball =8000kgm
terminal velocity(v)=25m/s^2
density of air( d) =1.29 kgm
now
volume of ball(V)=4/3pir^3=1.33×3.14×0.05^3=0.00052 m^3
density of ball= mass of ball/Volume of ball
or, 8000=m/0.00052
or, m=4.16 kg
weight of the ball (W)= mg=4.16×10=41.6 N
viscous force(F)=6 × pi × eta × r × v
=6×3.14×eta×0.05×25
=23.55×eta
To attain the terminal velocity,
Fiscous force=Weight
or, 23.55× eta = 41.6
or, eta = 1.76
whete eta is the coefficient of viscosity.
Answer:
17. NADH has a molar extinction coefficient of 6200 M2 cm at 340 nm. Calculate the molar concentration of NADH required to obtain an absorbance of 0.1 at 340 nm in a 1-cm path length cuvette. 18. A sample with a path length of 1 cm absorbs 99.0% of the incident light at a wavelength of 274 nm, measured with respect to an appropriate solvent blank. Tyrosine is known to be the only chromophore present in the sample that has significant absorption at 274 nm. Calculate the molar concentration of tyrosine in the sample.
Explanation:
