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andrey2020 [161]
4 years ago
14

Your friend of mass 150kg can just barely float in fresh water .Calculate her approximate volume. Express your answer to two sig

nificant figures and include the appropriate units.
Physics
1 answer:
zhuklara [117]4 years ago
4 0
The mind boggles at the spectacle ... a 150-kg lady weighs about 330 pounds, and has the courage and the audacity to not only go to the beach, but to go out and float in the water, hoping that there are no wise little kids around, getting ready to yell to each other "Thar she blows, Cap'n !".

But we're only here to do the math, whether or not it has any physical significance in the real world.

This problem is really not that mysterious. 

Since the star of this unlikely scenario is just barely floating, her weight must be equal to the weight of an equal volume of water.  In other words, her average density  (arrghhh, I can't take much more of this) is equal to the density of water.

Officially, 1 kg is the mass of fresh water that occupies a volume of 1 liter, so this relationship also very nearly describes the relationship between our heroine's mass and volume.

Her volume is very nearly 150 liters, or  0.15 cubic meter, and no further rounding is necessary, in more ways than one. 

I would stay and convert that to some more familiar 'customary' units for you, honest I would.  But I've been quite overcome by my visualizations of this whole matter, and I really must make a quick trip to the rail, immediately.
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3 years ago
In a double-slit interference experiment, which of the following actions (perhaps more than one) would cause the fringe spacing
frozen [14]

Answer:

The correct option is a , c

Explanation:

Generally the  fringe spacing is mathematically defined as

        y  =  \frac{m \lambda D}{d}

Where y is the fringe spacing

           m is the order of the fringe

           \lambda is the wavelength

           D is the distance between the slit and the screen

           d is the distance between the slit

Now in order to increase the fringe spacing can do the following

  • Increase the wavelength
  • increase the distance from the slit to the screen
  • Decrease the distance between the slit          

The following action would increase the fringe spacing because the from the question these parameters are directly proportional to the fringe spacing

3 0
3 years ago
A child has an ear canal that is 1.3 cm long. At what sound frequencies in the audible range will the child have increased heari
ElenaW [278]

Answer:

The answer to this is

6600 Hz to 19,800 Hz

Explanation:

The shape of the human ear is analogous to a closed ended pipe hence

we have λ = 4L or wavelength = 4 * length of the child ear

The frequency  c/λ  where c = speed of sound = 343 m/s

hence the child's audible range is multiples of 343/(4*0.13) =6600Hz

or 13200 Hz or 19,800 Hz

The generally quoted range of human hearing is 20 Hz to 20 kHz

7 0
3 years ago
In the far future, astronauts travel to the planet Saturn and land on Mimas, one of its 62 moons. Mimas is small compared with t
Mamont248 [21]

Answer:

a)h_{max}=14536.16 m

b)h = 15687.9 m

c)PD=7.62\% The estimate is low.

Explanation:

a) Using the energy conservation we have:

E_{initial}=E_{final}

we have kinetic energy intially and gravitational potential energy at the maximum height.

\frac{1}{2}mv^{2}=mgh_{max}

h_{max}=\frac{v^{2}}{2g}

h_{max}=\frac{43^{2}}{2*0.0636}

h_{max}=14536.16 m  

b)  We can use the equation of the gravitational force

F=G\frac{mM}{R^{2}}   (1)

We have that:

F = ma    (2)

at the surface G will be:

G=\frac{gR^{2}}{M}

Now the equation of an object at a distance x from the surface.

is:

F=\frac{mgR^{2}}{(R+x)^{2}}

m\frac{dv}{dt}=\frac{mgR^{2}}{(R+x)^{2}}

Using that dv/dt is vdx/dt and integrating in both sides we have:

v_{0}=\sqrt{\frac{2gRh}{R+h}}

h=\frac{v_{0}^{2}R}{2gR-v_{0}^{2}}

h=15687.9

c) The difference is:

So the percent difference will be:

PD=|\frac{14536.16-15687.9}{(14536.16+15687.9)/2}*100%

PD=7.62\%

The estimate is low.

I hope it helps you!

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