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romanna [79]
3 years ago
6

Two cylinders with the same mass density rC = 713 kg / m3 are floating in a container of water (with mass density rW = 1025 kg /

m3). Cylinder #1 has a length of L1 = 20 cm and radius r1 = 5 cm. Cylinder #2 has a length of L2 = 10 cm and radius r2 = 10 cm. If h1 and h2 are the heights that these cylinders stick out above the water, what is the ratio of the height of Cylinder #2 above the water to the height of Cylinder #1 above the water (h2 / h1)?
Physics
1 answer:
cluponka [151]3 years ago
3 0

Answer:

\dfrac{h_2}{h_1}=\dfrac{1}{2}

Explanation:

Lets take h is height of cylinder immersed in the water

We know that for floating body

h=\dfrac{\rho_cL}{\rho_l}

Where \rho_c density of cylinder

\rho_l density of water

For both cylinder fluid is same also density of cylinders are also same

So

\dfrac{h_1}{L_1}=\dfrac{h_2}{L_2}

\dfrac{h_1}{h_2}=\dfrac{L_1}{L_2}

\dfrac{h_1}{h_2}=\dfrac{20}{10}

\dfrac{h_2}{h_1}=\dfrac{1}{2}

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Answer:

The Troposphere is the lowermost portion of Earth’s atmosphere. It is the densest layer of the atmosphere and contains approximately 75 percent of the mass of the atmosphere and almost all the water vapor and aerosol. The troposphere extends from the Earth’s surface up to the tropopause where the stratosphere begins.

Explanation:

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3 years ago
A 0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of co
Genrish500 [490]

Answer: When 1.0kg of aluminium block is used, the final temperature of the mixture will be T = 36.2∘C

If 1.0kg copper block is used, T of the mixture will be = 17.4∘C

If 100g (0.1kg) of ice at 0∘C is used, T will be = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be= 147.1∘C

Explanation:

H = mcΘ

heat lost by block = heat gained by water

m₁c₁Θ₁ = m₂c₂Θ₂ where m₁ is mass of aluminium, m₂ is mass of water, c₁ is cAluminium, c₂ is cWater, Θ₁ is temperature change for aluminium, Θ₂ is temperature change for water.

0.5*900*(200-20) = m₁*4186*(20-0)

m₁ = 450*180/83270

<em>m₁ = 0.973kg</em>

<em>when 1.0kg of aluminium block is used, the final temperature of the mixture will be </em><em>T</em>

heat lost by block = heat gained by water

1.0*900*(200-T) = 0.973*4186*(T-0)

180000 - 900T = 4073T

4973T = 180000

T = 180000/4973 = 36.2∘C

<em>If 1.0kg copper block is used, T of the mixture will be</em>

heat lost by block = heat gained by water

1.0*387*(200-T) = 0.973*4186*(T-0)

77400 - 387T = 4073T

4460T = 77400

T = 77400/4460 = 17.4∘C

<em>If 100g (0.1kg) of ice at 0∘C is used, T will be</em>

<em>heat lost by block = heat gained by water + heat used in melting ice to form water at 0∘C</em>

heat used in melting 0.1kg of ice, H = ml, where l= 33600J/Kg

0.5*900*(200-T) = 0.1*4186*(T-0) + 0.1*33600J/Kg

90000 - 450T =  418.6T + 33600

418.6T + 450T = 90000 - 33600

868.6T = 56400

T = 56400/868.6 = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be

0.5*900*(200-T) = 0.025*4186*(T-0) + 0.025*33600J/Kg

90000 - 450T =  104.65T + 8400

104.65T + 450T = 90000 - 8400

554.65T = 81600

T = 81600/554.65 = 147.1∘C

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Read 2 more answers
A basketball player throws the ball at a 47 angle above the horizontal to a hoop which is located a horizontal distance L = 5.0
FromTheMoon [43]

Answer:

v_0 =1.71

Explanation:

the parabolic movment is described by the following equation:

y = tan(a)x-\frac{1}{2v_0^2(cos(a))^2}gx^2

where y is the height of the ball, a is the angle of launch, v_0 the initial velocity, g the gravity and x is the horizontal distance of the ball.

So, if we want that the ball reach the hood, we will replace values on the equation as:

0.8 = tan(47)(5)-\frac{1}{2v_0^2(cos(47))^2}(9.8)(5)^2

Finally, solving for v_0, we get:

v_0=\sqrt{\frac{-9.8(5)^2}{(0.8-tan(47)(5))2cos^2(47)}}

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4 0
4 years ago
A skier with a mass of 63 kg starts from rest and skis down an icy (frictionless) slope that has a length of 50 m at an angleof
victus00 [196]

Answer:

Explanation:

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h = 50 sin32 m

= 26.5 m

potential energy lost

= mgh

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mgh = 1/2 mv²

v = √ 2gh

= √ (2x9.8 x 26.5)

= 22.8 m /s

b )

Let μ be the coefficient of kinetic friction required.

friction force acting

= μmg

work done by friction in displacement of  d (40 m ) on horizontal surface

- μmg x d

This negative work will be equal to positive kinetic energy of the skier on horizontal surface .

=  μmg x d  = (1/2) m v²

μ = v² / (2 gd)

= 519.4 / (2 x 9.8 x 140 )

= .19

8 0
4 years ago
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