Answer:
500 N
Explanation:
Natural bouyency will keep the diver in the same place, no matter what the mass of gfs. This means if the diver is displacing with a weight of 500 N, the upthrust will also be 500 N.
Answer with Explanation:
We are given that
Mass of box=35 kg
Coefficient of static friction between box and truck bed=0.202
Acceleration due to gravity=
a.We have to find the force by which the box accelerates forward.
Force by which box accelerates=
Force by which box accelerates=
b.We have to find the maximum acceleration can the truck have before the box slides.
Force =friction force
Hence, the truck can have maximum acceleration before the box slide=
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.
Answer:
Approximate atomic radius for polonium-209 is 167.5 pm .
Explanation:
Number of atom in simple cubic unit cell = Z = 1
Density of platinum = 
Edge length of cubic unit cell= a = ?
Atomic mass of Po (M) = 209 g/mol
Formula used :
where,
ρ = density
Z = number of atom in unit cell
M = atomic mass
= Avogadro's number
a = edge length of unit cell
On substituting all the given values , we will get the value of 'a'.
Atomic radius of the polonium in unit cell = r
r = 0.5a
Approximate atomic radius for polonium-209 is 167.5 pm.
