How many does the stor6 need

A 12-kg piece of metal displaces 1.6 L of water when submerged. Part A Find its density. Express your answer to two significant

Tatiana [17]

Answer:

ρ = 7500 kg/m³

Explanation:

Given that

mass ,m = 12 kg

Displace volume ,V= 1.6 L

We know that

1000 m ³ = 1 L

Therefore V= 0.0016 m ³

When metal piece is fully submerged

We know that

mass = Density x volume

$m=\rho \times V$

Now by putting the values in the above equation

$\rho=\dfrac{12}{0.0016}\ kg/m^3$

ρ = 7500 kg/m³

Therefore the density of the metal piece will be 7500 kg/m³.

6 0

3 years ago

Can anyone explain how tides work

Mila [183]

The position of the sun and the moon affect how high the tide is

6 0

3 years ago

Read 2 more answers

2. A hanging wind-chime on a calm day would have kinetic or potential energy?

rjkz [21]

Answer:

it would have potential energy

7 0

3 years ago

What is a net force on an object that has a mass of 20.0 kg, an applied force of 100 n moving on a surface with a friction coeff

sergiy2304 [10]

The net force on the object as described is; 58.84N

Two forces acting on the object are;

The <em>applied force and the frictional force.</em>

In essence; the frictional force can be evaluated as;

Frictional force; = coefficient × Weight of object.

Frictional force = 0.21 × 20 × 9.8.

Frictional force = 41.16N

The Net force = Applied force - frictional force

Net force = 100 - 41.16N

Net Force = 58.84 N.

Read more:

brainly.com/question/94428

5 0

2 years ago

to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to

givi [52]

Answer:

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

$f_c = 10 Hz$ represent the corner frequency

$f= 60 Hz$ represent the original frequency

n represent the filter order and that's the variable that we need to find

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0

3 years ago