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Natasha2012 [34]

3 years ago

5

Two solid spheres are made from the same material, but one has twice the diameter of the other. Which sphere will have the great

er bulk modulus?

1 answer:

aivan3 [116]3 years ago

5 0

Answer:

It will be the same for both

Explanation:

from this question we have one similarity between these two spheres.

- they are both made from the same material,

The difference between both spheres is that:

- one of the spheres has its diameter to be twice as large as that of the other one.

We are to say the sphere with the greater bulk modulus.

If the material is the same thenthe Bulk modulus is also the same. It is not dependent on the material since it is a constant for that materia

Therefore the correct answer is:

It will be the same for both spheres.

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What is the most important element they keep in mind for the missions? of apollo

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Using your best estimates, how many times would you have to slap a 1 kg rotisserie chicken in order to cook it? You can assume t

FromTheMoon [43]

Answer:

n= 16021.03 slaps

Explanation:

Using law of Energy conservation

E_{thermal}= Kinetic energy of hand

⇒ $mc\Delta T= n\frac{1}{2}m_hv_h^2$

m_h= mass of the hand = 0.4 kg

v_h= velocity of the hand = 10 m/s

n= number of slaps

c= 4180 J/Kg °C

m= mass of chicken = 1 kg

Assuming all the energy of hand goes into chicken

Given Ti=0°C and T_f= 170 F= 76.66°C

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In the analogy for how a wave travels along a line of dominoes standing on end, each domino represents___________.

Alex777 [14]

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3 years ago

magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,

Mama L [17]

Answer:

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

Explanation:

Given:

height above which the rock is thrown up, $\Delta h=50\ m$

initial velocity of projection, $u=20\ m.s^{-1}$

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

$v=u+g'.t'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0=20-g'.t'$ (-ve sign to indicate that acceleration acts opposite to the velocity)

$t'=\frac{20}{g'}\ s$

The time taken by the rock to reach the top height on the earth:

$v=u+g.t$

$0=20-g.t$

$t=\frac{20}{g} \ s$

Height reached by the rock above the point of throwing on the exoplanet:

$v^2=u^2+2g'.h'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0^2=20^2-2\times g'.h'$

$h'=\frac{200}{g'}\ m$

Height reached by the rock above the point of throwing on the earth:

$v^2=u^2+2g.h$

$0^2=20^2-2g.h$

$h=\frac{200}{g}\ m$

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

$(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2$ (during falling it falls below the cliff)

here:

$u=$ initial velocity= 0 $m.s^{-1}$

$\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2$

$t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}$

$t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'} }$

Similarly on earth:

$t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g} }$

Now the required time difference:

$\Delta t=(t'+t_f')-(t+t_f)$

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

3 0

3 years ago