Pls answer it plssss

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 f

VashaNatasha [74]

Answer:

$\frac{y}{L}$ = 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

8 0

3 years ago

You’ve had practice calculating the grams of hydrogen gas, but it is also possible to calculate the amount of oxygen gas produce

alekssr [168]

Answer:41.991ml

Explanation:

Equations: 2 H2O → 4H+ + 4e + O2 OXIDATION

2 H+ + 2e → H2 REDUCTION

Electrolysis is the chemical decomposition of compounds when electricity is made to pass through a molten compound or solution.

from the oxidation reaction:

1moles of oxygen requires 4moles of electrons to be discharged at the product

F=96500C/mol

Quantity of charge Q=It

=60*60*0.201A

Q=723.6C

Mole=Q/(F*mole ratio of electron)

Mole= 723.6/(4*96500)

Mole=((1809)/(965000))

M=0.0018746114

M1/M2=V1/V2

1/0.00187=22.4dm^3/V2

V2=22.4*0.00187

V2=0.04199129534dm^3

41.99129534ml

5 0

2 years ago

A drunken sailor stumbles 580 meters north, 530 meters northeast, then 480 meters northwest. What is the total displacement and

kvv77 [185]

Answer:

(a) 1294.66 m

(b) 88.44°

Explanation:

d1 = 580 m North

d2 = 530 m North east

d3 = 480 m North west

(a) Write the displacements in vector forms

$\overrightarrow{d_{1}}=580\widehat{j}$

$\overrightarrow{d_{2}}=530\left ( Cos45\widehat{i}+Sin45\widehat{j} \right )$

$\overrightarrow{d_{2}}=374.77\widehat{i}+374.77\widehat{j}$

$\overrightarrow{d_{3}}=480\left ( - Cos45\widehat{i}+Sin45\widehat{j} \right )$

$\overrightarrow{d_{3}}=-339.41\widehat{i}+339.41widehat{j}$

The resultant displacement is given by

$\overrightarrow{d}\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}$

$\overrightarrow{d}=\left ( 374.77-339.41 \right )\widehat{i}+\left ( 580+374.77+339.41 \right )\widehat{j}$

$\overrightarrow{d}=35.36\widehat{i}+1294.18\widehat{j}$

magnitude of the displacement

$d ={\sqrt{35.36^{2}+1294.18^{2}}}=1294.66 m$

d = 1294.66 m

(b) Let θ be the angle from + X axis direction in counter clockwise

$tan\theta =\frac{1294.18}{35.36}=36.6$

θ = 88.44°

4 0

3 years ago

A child with a weight of 430 N rides on a Ferris wheel, which has a radius of 17 m, and the linear velocity of the 3.5 m/s at an

Murrr4er [49]

At the lowest point on the Ferris wheel, there are two forces acting on the child: their weight of 430 N, and an upward centripetal/normal force with magnitude n; then the net force on the child is

∑ F = ma

n - 430 N = (430 N)/g • a

where m is the child's mass and a is their centripetal acceleration. The child has a linear speed of 3.5 m/s at any point along the path of the wheel whose radius is 17 m, so the centripetal acceleration is

a = (3.5 m/s)² / (17 m) ≈ 0.72 m/s²

and so

n = 430 N + (430 N)/g (0.72 m/s²) ≈ 460 N

6 0

2 years ago

If you have 12 atoms of hydrogen before a chemical reaction how many atoms of hydrogen will be president after the chemical reac

kaheart [24]

If you have 12 atoms of hydrogen before a chemical reaction, the number of hydrogen atoms that will be present after the chemical reaction is 12 atoms.
The Law of Conservation of Mass (LOCOM) states that mass is neither created nor destroyed before and after any chemical reaction.
According to the Law of Conservation of Mass (LOCOM), a balanced chemical equation requires that the number of atoms on the reactant side must be equal to the number of atoms on the product side of any chemical reaction.
In this context, a chemical reaction having 12 atoms of hydrogen as reactants at the beginning, should also produce a total of 12 atoms of hydrogen as products at the end of the chemical reaction.

4 0

2 years ago