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3 years ago

What volume of water can the aquarium hold?

Mathematics

2 answers:

natta225 [31]3 years ago

5 0

Answer: The volume of water that can be hold by aquarium is 46200 cm³

Step-by-step explanation:

Since, the volume of a cuboid is,

V = l × b × h

Where, l = length, b = width and h = height.

Here, l = 50 cm, b = 28 cm and h = 33 cm,

Thus, the volume of the water tank,

V = 50 × 28 × 33 = 46200 cm³

Hence, the volume of water that can be hold by aquarium is 46200 cm³

Anni [7]3 years ago

4 0

Formula is l x w x h
50x28x33=46200 cm^3
last choice 46,200 cm^3

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Svet_ta [14]

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4 years ago

A baseball player on a professional baseball team has 207 hits in 575 at-bats. What

Vika [28.1K]

Answer:

9/25

Step-by-step explanation:

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2 years ago

Round 392,153 to the nearest hundred thousand

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3 years ago

Manny and Evan were flying paper airplanes. Manny’s airplane flew 36.5 centimeters and Evan’s airplane flew 18 centimeters. How

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3 years ago

Read 2 more answers

A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and

nasty-shy [4]

Answer:

correct option is C) 2.8

Step-by-step explanation:

given data

string vibrates form = 8 loops

in water loop formed = 10 loops

solution

we consider mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = $\rho$

case (1)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

so here

$l = \frac{8 \lambda _1}{2}$ ..............1

$l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}$

and we know velocity is express as

velocity = frequency × wavelength .....................2

$\sqrt{\frac{Tension}{mass\ per\ unit \length }}$ = f × $\lambda_1$

here tension = mg

$\sqrt{\frac{mg}{\mu}}$ = f × $\lambda_1$ ..........................3

and

case (2)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

$l = \frac{10 \lambda _1}{2}$ ..............4

$l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}$

when block is immersed

equilibrium eq will be

Tenion + force of buoyancy = mg

T + v × $\rho$ × g = mg

and

T = v × $\rho$ - v × $\rho$ × g

from equation 2

f × $\lambda_2$ = f × $\frac{1}{5}$

$\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}$ .......................5

now we divide eq 5 by the eq 3

$\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}$

solve irt we get

$1 - \frac{\rho _{stone}}{\rho _{water}} = \frac{16}{25}$

relative density $\frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}$

relative density = 2.78 ≈ 2.8

so correct option is C) 2.8

3 0

4 years ago