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prisoha [69]
3 years ago
11

A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate

r at the bottom of the well 5.00 seconds after being dropped from the tower. Calculate the depth of the well. Given: g = -9.81 meters/second2.
Physics
1 answer:
ELEN [110]3 years ago
7 0

A useful formula that gives the free-fall distance from rest in 'T' seconds:

                       D = (1/2 G) x (T²)

      G = 9.81 m/s²
1/2 G = 4.905 m/s²

                       D (5 seconds) = (4.905 m/s²) x (5 sec)²

                                              = (4.905 m/s²) x (25 sec²)

                                              =        122.625 meters .

Since the tower-top is 100m above ground,
the depth of the well, to the top of the water,
accounts for the additional  22.625 meters.

My question is:  How do you know exactly when the stone hit the water ?

You probably stood at the top of the well and listened for the
sound of the 'plop'.  But it took some time after the stone hit
the water for the sound of the plop to come back up to you. 
Well, can't you just subtract that time ?  Yes, but you need
to know how much time to subtract.  That depends on the
depth of the well ... which is exactly what you're trying to
determine, so you don't know it yet.

Oh well.  That's a deep subject.
 
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anyanavicka [17]

Answer:

I = 1.5*10⁻³ kg*m²

Explanation:

  • It can be showed that the moment of inertia (or rotational inertia) for a uniform cylinder of mass m and radius r, respect an longitudinal axis going through its center (parallel to the height of the cylinder) can be written as follows:

       I = \frac{1}{2}*m*r^{2}  = \frac{1}{2}*0.400 kg*(0.0865m)^{2}  = 1.5e-3 kg*m2

3 0
3 years ago
An object in the shape of a thin ring has radius a and mass M. A uniform sphere with mass m and radius R is placed with its cent
madreJ [45]

Answer:

F = GMmx/[√(a² + x²)]³

Explanation:

The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is

dF = GmdM/L²

Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.

So, the horizontal components add from two symmetrically opposite mass elements dM,

Thus, the horizontal component of the force is

dF' = dFcosФ where Ф is the angle between L and the x axis

dF' = GmdMcosФ/L²

L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.

L = √(a² + x²)

cosФ = x/L

dF' = GmdMcosФ/L²

dF' = GmdMx/L³

dF' = GmdMx/[√(a² + x²)]³

Integrating both sides we have

∫dF' = ∫GmdMx/[√(a² + x²)]³

∫dF' = Gm∫dMx/[√(a² + x²)]³    ∫dM = M

F = GmMx/[√(a² + x²)]³  

F = GMmx/[√(a² + x²)]³

So, the force due to the sphere of mass m is

F = GMmx/[√(a² + x²)]³

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The basic principles that apply to circuits will be;

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2. Electrons transfer energy to perform some useful function.

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