Answer:
If a car is moving at a <em><u>constant velocity of 10 m/s,</u></em> there will be no change in the velocity per time.
Acceleration is the rate of change of velocity
Acceleration=(Final velocity-Initial velocity)/Time
Acceleration=(10m/s-10m/s)/Time= <em><u>0 m/s²</u></em>
<h3>★ <u>0 m/s²</u> is the right answer. </h3>
Answer:
F = 4.147 × 10^23
v = 1.31 × 10^4
Explanation:
Given the following :
mass of Jupiter (m1) = 1.9 × 10^27
Mass of sun (m2) = 1.99 × 10^30
Distance between sun and jupiter (r) = 7.8 × 10^11m
Gravitational force (F) :
(Gm1m2) / r^2
Where ; G = 6.673×10^-11 ( Gravitational constant)
F = [(6.673×10^-11) × (1.9 × 10^27) × (1.99 × 10^30)] / (7.8 × 10^11)^2
F = [25.231 × 10^(-11+27+30)] / (60.84 × 10^22)
F = (25.231 × 10^46) / (60.84 × 10^22)
F = 3.235 × 10^(46 - 22)
F = 0.4147 × 10^24
F = 4.147 × 10^23
Speed of Jupiter (v) :
v = √(Fr) / m1
v = √[(4.147 × 10^23) × (7.8 × 10^11) / (1.9 × 10^27)
v = √32.3466 × 10^(23+11) / 1.9 × 10^27
v = √32.3466× 10^34 / 1.9 × 10^27
v = √17. 023 × 10^34-27
v = √17.023 × 10^7
v = 13047.221
v = 1.31 × 10^4
The gravity is pushing rhe boat down
According to Charles law, we know, at constant pressure, volume is directly proportional to temperature.
So, <span>V/T = constant
</span>
V₁/t₁ = V₂/t₂
V₁t₂ = V₂t₁
Here, we have: V₁ = 9 mL
V₂ = ?
T₂ = 50+272 = 323 K
T₁ = 19+273 = 292 K
Substitute their values into the expression:
9 × 323 = V₂ × 292
V₂ = 2907 / 292
V₂ = 9.95
After rounding-off to unit place value, it would be equal to 10 mL
So, In short Option C would be your correct answer.
Hope this helps!
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.
