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2 years ago

7

What fraction of a piece of concrete will be submerged when it floats in mercury? the density of concrete is 2.3×103kg/m3 and de

nsity of mercury is 13.6×103kg/m3?

1 answer:

chubhunter [2.5K]2 years ago

3 0

<span>17% Any object that is floating will display the volume of fluid that matches the mass of the object that's floating. So if the object is 100% as dense as the fluid, it will just barely sink, if it's 50% as dense as the fluid, 50% will be submerged, etc. So let's see what percent of mercury's density is concrete. 2.3x10^3 / 13.6x10^3 = 0.169117647 = 16.9117647% Rounding to 2 significant figures gives 17%. So 17% of a piece of concrete will be submerged in mercury when it's floating in mercury.</span>

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Answer:

Explanation:

Given

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apparent Frequency $f'=697\ Hz$

There is change in frequency whenever source move relative to the observer.

From Doppler effect we can write as

$f'=f\cdot \frac{v-v_o}{v+v_s}$

where

$f'=$ apparent frequency

v=velocity of sound in the given media

$v_s=$ velocity of source

$v_0=$ velocity of observer

here $v_0=0$

$697=723\cdot (\frac{343-0}{343+v_s})$

$v_s=(\frac{f}{f'}-1)v$

$v_s=(\frac{723}{697}-1)\cdot 343$

$v_s=12.79\approx 12.8\ m/s$

i.e.fork acquired a velocity of $12.8 m/s$

distance traveled by fork is given by

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v_s^2-0=2\times 9.8\times s$

$s=\frac{12.8^2}{2\times 9.8}$

$s=8.35\ m$

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2 years ago

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Answer:

3311N

Explanation:

r = radius = 600m

V = speed = 150m/s

Mass = weight = 70kg

The weight of pilot when calculated due to circular motion

W = tv

Fv = mv²/r

Fv = 70x150²/600

Fv = 79x22500/600

= 15750000/600

= 2625N

Real Weight of the pilot = m x g

= 70 x 9.8

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The apparent Weight is calculated by

Mv²/r + mg

= 2625N + 686N

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Therefore the apparent Weight is 3311N

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valentina_108 [34]

Answer:

3. none of these

Explanation:

The rotational kinetic energy of an object is given by:

$K=\frac{1}{2}I \omega^2$

where

I is the moment of inertia

$\omega$ is the angular speed

In this problem, we have two objects rotating, so the total rotational kinetic energy will be the sum of the rotational energies of each object.

For disk 1:

$K_1 = \frac{1}{2}I (\omega)^2 = \frac{1}{2}I\omega^2$

For disk 2:

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so the total energy is

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Answer:

Please see attachment

Explanation:

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2 years ago