Answer:
The correct option is
a. v = 
Explanation:
Time at which the object start fall t = 0
The acceleration a is given by a = g - bV
Where V = Speed of the object
Speed V² = u² + 2·a·h
However with the drag force the object will approach terminal velocity as t becomes progressively larger whereby v will stop increasing
Option a. is the only option that has limiting value of v which is in the range of g as t increases ∴ option a. is the correct option.
v = as t increases 
→ 1 s and v→ g/b m/s
Answer:
1.6 m/s
Explanation:
First you need to find the momentums of each disc by multiplying their velocities with mass.
disc 1: 7*1= 7 kg m/s
disc 2: 1*9= 9 kg m/s
Second, you need to find the total momentum of the system by adding the momentums of each sphere.
9+7= 16 kg m/s
Because momentum is conserved, this is equal to the momentum of the composite body.
Finally, to find the composite body's velocity, divide its total momentum by its mass. This is because mass*velocity=momentum
16/10=1.6
The velocity of the composite body is 1.6 m/s.
The number of significant digits to the answer of the following problem is four.
<h3>What are the significant digits?</h3>
The number of digits rounded to the approximate integer values are called the significant digits.
The following problem is
(2.49303 g) * (2.59 g) / (7.492 g) =
On solving we get
= 0.86184566204
The answer is approximated to 0.86185
Thus, the significant digits must be four.
Learn more about significant digits.
brainly.com/question/1658998
#SPJ1
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.
