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Brrunno [24]
3 years ago
15

A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber

, and from the bottom hangs a 10 gram lead weight. The density of lead is 11.3 g/cm3. What fraction of the bobber's volume is submerged, as a percent of the total volume
Physics
1 answer:
Tasya [4]3 years ago
7 0

Answer:

Explanation:

Calculate the volume of the lead

V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}

Now calculate the bouyant force acting on the lead

F_L = Vpg

F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N

This force will act in upward direction

Gravitational force on the lead due to its mass  will act in downward direction

Hence the difference of this two force

T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N

If V is the volume submerged in the water then bouyant force on the bobber is

F_B=V'pg

Equate bouyant force with the tension and gravitational force

F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}

Now Total volume of bobble is

\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\

=\large\boxed{4.52 \%}

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A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision
marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

x \approx 103,46x10^{9}m.

B) The space time coordinate t of the collision in Earth's reference frame is

t=377,29s

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

                          x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                                y'=y

                                z'=z

                          t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

                       x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                           y=y'

                           z=z'

                        t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

First we calculate the expression in the denominator

                            \frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}

                                \sqrt{1-\frac{v^{2}}{c^{2}}} =0,714

then we calculate t

                      t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                      t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}

                      t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}

                      t=\frac{190s+79,388s}{0,714}

finally we get that

                                     t=377,29s

then we calculate x

                         x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                         x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+39872396914m}{0,714}}

                         x=\frac{73872396914m}{0,714}}

                         x=103462740775,91m

finally we get that

                                     x \approx 103,46x10^{9} m

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lorasvet [3.4K]
Drop "moves" from the list for a moment.

You can also drop "stops moving", because that's included in "changes speed"
(from something to zero).

When an object changes speed or changes direction, that's called "acceleration".

I dropped the first one from the list, because an object can be moving,
and as long as it's speed is constant and it's moving in a straight line,
there's no acceleration.

I think you meant to say "starts moving". That's a change of speed (from zero
to something), so it's also acceleration.
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