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vladimir2022 [97]
1 year ago
14

when a metal sphere is dropped in to a tall cylinder containing liquid its acceleration is g÷2 (gravity over 2) show that : dens

ity of metal =2density of liquid​
Physics
1 answer:
Kryger [21]1 year ago
6 0

The density of the metal sphere is 2 times the density of the liquid as proved.

<h3>Net upward force acting on the metal sphere</h3>

The net upward force acting on the sphere as it is dropped into the liquid is calculated as follows;

F =  σVg - ρVg

ma =  σVg - ρVg  

where;

  • ρ is density of the liquid
  • σ is the density of the metal
  • a is acceleration of the metal

σV(a) =  σVg  - ρVg

σ(a) = σg  - ρg

σ(g/2) = σg -  ρg  

g(σ/2) = g(σ -  ρ)

σ/2 = σ -  ρ

σ/2 - σ = -  ρ

-σ/2 = -  ρ

σ = 2ρ

Thus, the density of the metal sphere is 2 times the density of the liquid as proved.

Learn more about density here: brainly.com/question/1354972
#SPJ1

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A gamma ray photon has an energy of 0.91 GeV. (1 GeV = 109 eV.) What is the wavelength of the gamma ray in fm? (1 fm = 10-15 m)?
anastassius [24]

Answer:

\lambda=1.37 fm

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6 0
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During the latter part of your European vacation, you are hanging out at the beach at the gold coast of Spain. As you are laying
Jlenok [28]

Well, I guess you can come close, but you can't tell exactly.

It must be presumed that the seagull was flying through the air
when it "let fly" so to speak, so the jettisoned load of ballast
of which the bird unburdened itself had some initial horizontal
velocity.

That impact velocity of 98.5 m/s is actually the resultant of
the horizontal component ... unchanged since the package
was dispatched ... and the vertical component, which grew
all the way down in accordance with the behavior of gravity.

  98.5 m/s  =  √ [ (horizontal component)² + (vertical component)² ].

The vertical component is easy; that's (9.8 m/s²) x (drop time).
Since we're looking for the altitude of launch, we can use the
formula for 'free-fall distance' as a function of acceleration and
time:

             Height = (1/2) (acceleration) (time²) .

If the impact velocity were comprised solely of its vertical
component, then the solution to the problem would be a
piece-o-cake.

                  Time = (98.5 m/s) / (9.81 m/s²) = 10.04 seconds
whence
                 Height = (1/2) (9.81) (10.04)²

                            =   (4.905 m/s²) x (100.8 sec²)  =  494.43 meters.

As noted, this solution applies only if the gull were hovering with
no horizontal velocity, taking careful aim, and with malice in its
primitive brain, launching a remote attack on the rich American.

If the gull was flying at the time ... a reasonable assumption ... then
some part of the impact velocity was a horizontal component.  That
implies that the vertical component is something less than 98.5 m/s,
and that the attack was launched from an altitude less than 494 m.   

8 0
3 years ago
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