You asked a question. I'm about to answer it.
Sadly, I can almost guarantee that you won't understand the solution.
This realization grieves me, but there is little I can do to change it.
My explanation will be the best of which I'm capable.
Here are the Physics facts I'll use in the solution:
-- "Apparent magnitude" means how bright the star appears to us.
-- "Absolute magnitude" means the how bright the star WOULD appear
if it were located 32.6 light years from us (10 parsecs).
-- A change of 5 magnitudes means a 100 times change in brightness,
so each magnitude means brightness is multiplied or divided by ⁵√100 .
That's about 2.512... .
-- Increasing magnitude means dimmer.
Decreasing magnitude means brighter.
+5 is 10 magnitudes dimmer than -5 .
-- Apparent brightness is inversely proportional to the square
of the distance from the source (just like gravity, sound, and
the force between charges).
That's all the Physics. The rest of the solution is just arithmetic.
____________________________________________________
-- The star in the question would appear M(-5) at a distance of
32.6 light years.
-- It actually appears as a M(+5). That's 10 magnitudes dimmer than M(-5),
because of being farther away than 32.6 light years.
-- 10 magnitudes dimmer is ( ⁵√100)⁻¹⁰ = (100)^(-2) .
-- But brightness varies as the inverse square of distance,
so that exponent is (negative double) the ratio of the distances,
and the actual distance to the star is
(32.6) · (100)^(1) light years
= (32.6) · (100) light years
= approx. 3,260 light years . (roughly 1,000 parsecs)
I'll have to confess that I haven't done one of these calculations
in over 50 years, and I'm not really that confident in my result.
If somebody's health or safety depended on it, or the success of
a space mission, then I'd be strongly recommending that you get
a second opinion.
But, quite frankly, I do feel that mine is worth the 5 points.
They hit at the same time. Everything falls at the same rate.
So first Identify all the given Varibales so u can choose which Eqauton to use
D=200m
T=4s
Vi=10m/s
Vf=?
You should this equation
D= 0.50(Vf+Vi)T
Plug in the values
200= 0.50 (Vf+10) 4
Divide the 4 out of the right side and if you do sumthing to one side you gotta do it to the other
200 divided by 4= 0.50(Vf+10)
50= 0.50(Vf+10)
Now expand the 0.50
So 50= 0.5Vf + 5 (because 0.5 times 10 is 5)
Now get rid of the 5
50-5= 0.5Vf
45 =0.5Vf now Divide the 0.5 out
45 divided by 0.5 = Vf
And 45/0.5 is 90
So 90=Vf
Therefore the final Velocity is 90m/s
Answer:
D. 21 ml
Explanation:
Since, the cylinder is marked and graduated in the intervals if 1 ml. Therefore, the values between two consecutive ml, such as between 30 ml and 31 ml can not be determined. Because, we do not have any scale in between the ml. So, the least count of this instrument is 1 ml. This graduated cylinder can give the answers to zero decimal places, accurately. And it can not determine any decimal value due to its graduating or the marking limitation. So, all the options given, contain a decimal value, except for the option D. In option D there is no decimal value, hence it is a correct answer.
D. <u>21 ml</u>
Using idea of conservation of impulse-momentum theorem, the instantaneous velocity times mass is equal to force times the change in time.
mv = Ft
To reduce the force, decrease the velocity and mass. You can also extend the time of the collision. This is why cars collapse on impact. They were design that way to reduce the force on the car and the passenger.
Hopes this helps!
