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2 years ago

7

A metal block suspended from a spring balance is submerged in water. You observe that the block displaces 55 cm3 of water and th

at the balance reads 4.3 N. What is the density of the block?

1 answer:

DiKsa [7]2 years ago

4 0

Answer:

8977.7 kg/m^3

Explanation:

Volume of water displaced = 55 cm^3 = 55 x 10^-6 m^3

Reading of balance when block is immersed in water = 4.3 N

According to the Archimedes principle, when a body is immersed n a liquid partly or wholly, then there is a loss in the weight of body which is called upthrust or buoyant force. this buoyant force is equal to the weight of liquid displaced by the body.

Buoyant force = weight of the water displaced by the block

Buoyant force = Volume of water displaced x density of water x g

= 55 x 10^-6 x 1000 x .8 = 0.539 N

True weight of the body = Weight of body in water + buoyant force

m g = 4.3 + 0.539 = 4.839

m = 0.4937 kg

Density of block = mass of block / volume of block

= $\frac{0.4937}{55\times10^{-6}}$

Density of block = 8977.7 kg/m^3

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3. A large crane lifts a 25,000 kg mass in the air. The amount of work that must be done by the

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$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the concept of Efficiency.

Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%

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so we get as,

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The period of rotation of Mars is 1 day and 37 minutes. Determine its frequency of rotation in Hertz.

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The frequency of rotation of Mars is 0.0000113 Hertz.

<u>Given the following data:</u>

Period = 1 day and 37 minutes.

To find the frequency of rotation in Hertz:

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Cross-multiplying, we have:

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Read more: brainly.com/question/14708169

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2 years ago

What is the pressure drop due to the Bernoulli Effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire

Marianna [84]

Answer:

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Given that,

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Using Bernoulli effect

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho g h_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

When two point are at same height so ,

$P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2$ ....(I)

Firstly we need to calculate the velocity

Using continuity equation

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$Q=A_{1}v_{1}$

$v_{1}=\dfrac{Q}{A_{1}}$

$v_{1}=\dfrac{40.0\times10^{-3}}{\pi\times(1.5\times10^{-2})^2}$

$v_{1}=56.58\ m/s$

For output velocity,

$v_{2}=\dfrac{40.0\times10^{-3}}{\pi\times(4.5\times10^{-2})^2}$

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(b). We need to calculate the maximum height

Using formula of height

$\Delta P=\rho g h$

Put the value into the formula

$1.58\times10^{6}=1000\times9.8\times h$

$h=\dfrac{1.58\times10^{6}}{1000\times9.8}$

$h=161.22\ m$

Hence, The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

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3 years ago