Physics, 07.01.2021 17:00 angel478

A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the fraction of the length of the rod above water

Answers: 3

Show answers

Answers

Answer from: arichar

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 = X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + $(\frac{L-y}{2})$ ) g sinФ - F2 x ( $\frac{L}{2}$ ) x gsinФ = 0

plug in the equations of F1 and F2 into the above equation and after simplification, we get:

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . 5/9 x w

Hence,

$(L^{2} - y^{2} )$ = $\frac{5L^{2} }{9}$

$\frac{L^{2} - y^{2} }{L^{2} }$ = $\frac{5}{9}$

Taking $L^{2}$ common and solving for $\frac{y}{L}$ , we will get

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

Answer from: Quest

found in the nucleus

has mass of one amu

Answer from: Quest

it breaks and dirupture other materials and other sencews from the bottom

Another question on Physics

Physics, 21.06.2019 18:20

What color would an object that reflects green and blue light appear to be in white light? a. blue b. cyan c. magenta d. black

Answers: 1

Answer

Physics, 22.06.2019 13:00

Discuss how the hardness or softness of the landing surface is related to the time required to stop the egg

Answers: 1

Answer

Physics, 22.06.2019 21:40

Wo small variable-thrust jets are actuated to keep the spacecraft angular velocity about the z-axis constant at ? 0 = 1.16 rad/s as the two telescoping booms are extended from r1 = 1.18 m to r2 = 4.69 m at a constant rate over a period of 124 seconds. the small 19-kg experiment modules at the ends of the booms may be treated as particles, and the mass of the rigid booms is negligible. determine the necessary thrust t for each jet as a function of time where t = 0 is the time when the telescoping action is begun. after you have the general expression for t, answer the questions. show work.

Answers: 1

Answer

Physics, 22.06.2019 22:30

Use the drop-down menus to answer each question. which line represents the movement of a cart with a constant acceleration? a which line represents the movement of a cart with a constant velocity? c which lines represents a cart with no velocity or acceleration? b

Answers: 1

Answer

You know the right answer?

A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5...

Questions