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Mice21 [21]
3 years ago
5

13 A cylindrical iron rod 8 cm high and 6 cm in diameter stands in a cylindrical tin 12 cm in diameter. See Fig. 15.22. Water is

poured into the tin until its depth is 8 cm. How far would the level drop when the rod is removed?​
Mathematics
2 answers:
gulaghasi [49]3 years ago
5 0
The correct answer is 2 cm
Eduardwww [97]3 years ago
4 0

Answer:

2cm

Step-by-step explanation:

The volume of water displaced by the rod is

v = πr^2 h = π * 3^2 * 8 = 72π cm^3

The drop in water level is the rod volume divided by the cross-section area of the tin, or

72π/(π*6^2) = 2 cm

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Solve for <br> 3x + 12= -12
oee [108]

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Step-by-step explanation:

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4 0
4 years ago
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In △ABC, AB = 13.2m,
luda_lava [24]

Answer:

(i) ∠ABH  = 14.5°

(ii) The length of AH = 4.6 m

Step-by-step explanation:

To solve the problem, we will follow the steps below;

(i)Finding  ∠ABH

first lets find <HBC

<BHC + <HBC + <BCH  = 180°  (Sum of interior angle in a polygon)

46° + <HBC  + 90 = 180°

 <HBC+ 136°  = 180°

subtract 136 from both-side of the equation

 <HBC+ 136° - 136°  = 180° -136°

 <HBC  = 44°

lets find <ABC

To do that, we need to first find <BAC

Using the sine rule

\frac{sin A}{a} =  \frac{sin C}{c}

A = ?

a=6.9

C=90

c=13.2

\frac{sin A}{6.9} = \frac{sin 90}{13.2}

sin A = 6.9 sin 90  /13.2

sinA = 0.522727

A = sin⁻¹ ( 0.522727)

A ≈ 31.5 °

<BAC  = 31.5°

<BAC + <ABC + <BCA = 180° (sum of interior angle of a triangle)

31.5° +<ABC + 90° = 180°

<ABC  + 121.5°  = 180°

subtract 121.5° from both-side of the equation

<ABC  + 121.5° - 121.5°  = 180° - 121.5°

<ABC = 58.5°

<ABH = <ABC - <HBC

           =58.5° - 44°

            =14.5°

∠ABH = 14.5°

(ii) Finding the length of AH

To find length AH, we need to first find ∠AHB

<AHB + <BHC = 180°  ( angle on a straight line)

<AHB + 46° = 180°

subtract 46° from both-side of the equation

<AHB + 46°- 46° = 180° - 46°

<AHB  = 134°

Using sine rule,

\frac{sin 134}{13.2}  = \frac{sin 14.5}{AH}

AH = 13.2 sin 14.5 / sin 134

AH≈4.6 m

length AH = 4.6 m

8 0
3 years ago
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