All categories

3 years ago

What % of an object’s mass is above the water line if the object’s density is 0.82g/ml?

1 answer:

4 0

To develop this problem we will apply the Archimedes model. As well as the definitions of Weight based on mass and acceleration. The first in turn will be considered under the relationship of Density and Volume. From the values given we have to:

$\rho_w =1 g/mL \rightarrow \text{Water Density}$

$\rho_o = 0.82g/mL \rightarrow \text{Object density}$

Since it is in equilibrium, the weight of the object will have a reaction from the water, which will cause the sum of forces between the two objects to be zero, therefore

$\sum F= 0$

$F_w-F_o = 0$

$F_w = F_o$

$m_w g = m_og$

$\rho_w V_w g = \rho_o V_o g$

The value of gravity is canceled because it is a constant

$\frac{V_w}{V_o} = \frac{\rho_o}{\rho_w}$

$\frac{V_w}{V_o} = \frac{0.82}{1}$

$\frac{V_w}{V_o} = 0.82$

The portion of the object that is submerged corresponds to 82%, while the portion that is visible, above the water level will be 18%

You might be interested in

If the star Sirius emits 23 times more energy than the Sun, why does the Sun appear brighter in the sky?

Ganezh [65]

Answer:

As b ∝ (L/r²) and

the distance of the sun from the earth is 0.00001581 light years

and

the distance of the Sirius from the earth is 8.6 light years

hence,

the Sun appear brighter in the sky

Explanation:

The brightness (b) is directly proportional to the Luminosity of the star (L) and inversely proportional to the square of the distance between the star and the observer (r).

thus, mathematically,

b ∝ (L/r²)

now,

given

L for sirius is 23 times more than the sun i.e 23L

now,

the distance of the sun from the earth is 0.00001581 light years

and

the distance of the Sirius from the earth is 8.6 light years

thus,

using the the relation between conclude that the value of brightness for the Sirius comes very very low as compared to the value for brightness for the Sun.

hence, the sun appears brighter

5 0

3 years ago

Vaccine ASAP don't got much time

djyliett [7]

Answer:

i answered this question not that long ago....it's D

Explanation:

7 0

3 years ago

Read 2 more answers

Find the ratio of the diameter of iron to copper wire, if they have the same resistance per unit length (as they might in househ

Natasha_Volkova [10]

Answer:

The ratio of the diameter of iron to Cu is;

$\frac{d{Fe} }{ d{Cu} } =\sqrt{\frac{p_{Fe} }{ p_{Cu} }}$

Explanation:

R=(ρL)/A

R is resistance,
L is length,
A is area,
ρ is resistivity
d is diameter

from the question the two materials have the same resistance per unit length.

$\frac{R}{L}= \frac{p}{A}$

$\frac{R}{L}$ for iron = $\frac{R}{L}$ for copper

This means we can equate ρ/A for both materials.

$\frac{p_{Fe} }{A_{Fe} } =\frac{p_{Cu} }{A_{Cu} }$

re-arranging the equation we have,

$\frac{A_{Fe}}{A_{Cu} } =\frac{p_{Fe} }{ p_{Cu} }$

$A=\pi \frac{d^{2} }{4}$

$\frac{A_{Fe}}{A_{Cu} } =\frac{d^{2}{Fe} }{ d^{2}{Cu} }$

$\frac{d^{2}{Fe} }{ d^{2}{Cu} } =\frac{p_{Fe} }{ p_{Cu} }$

$\frac{d{Fe} }{ d{Cu} } =\sqrt{\frac{p_{Fe} }{ p_{Cu} }}$

5 0

3 years ago

A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The

S_A_V [24]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

$\sum F_x = 0$

$f-N_w = 0$

$N_w = f$

The forces in the horizontal direction would be,

$\sum F_y = 0$

$N_f -W =0$

$N_f = W$

The sum of Torques at equilibrium,

$\sum \tau = 0$

$Wdcos\theta - N_wLsin\theta = 0$

$WdCos\theta = fLSin\theta$

$f = \frac{Wd}{Ltan\theta}$

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

$f_{max} = \mu W=\frac{Wd}{Ltan\theta}$