How much protein would be consumed with a 3-ounce serving of

A physicist comes across an open container which is filled with two liquids. Since the two liquids have different density, there

Dominik [7]

Answer:

496.57492 kg/m³

Explanation:

$P_a$ = Atmospheric pressure = 101300 Pa

$\rho_w$ = Density of water = 1000 kg/m^3

$h_w$ = Height of water = 21.8 cm

$h_f$ = Height of fluid = 30 cm

g = Acceleration due to gravity = 9.81 m/s²

$\rho_f$ = Density of the unknown fluid

Absolute pressure at the bottom

$P_{abs}=P_a+\rho_wgh_w+\rho_fgh_f\\\Rightarrow \rho_f=\frac{P_{abs}-P_a-\rho_wgh_w}{gh_f}\\\Rightarrow \rho_f=\frac{104900-101300-1000\times 9.81\times 0.218}{9.81\times 0.3}\\\Rightarrow \rho_f=496.57492\ kg/m^3$

The density of the unknown fluid is 496.57492 kg/m³

6 0

3 years ago

A closed cylindrical tank of radius 3.5 m and height 2m is made from

Dahasolnce [82]

Explanation:

Total surface area of cylinder:

$2\pi \times r(h + r) \\ 2 \times \pi \times 3.5(3.5 + 2) \\ 2\pi \times 19.25 = 120.951317 \\ 120.95 \: {cm}^{2}$

sheet of metal required = 120.95 cm^2

3 0

3 years ago

Read 2 more answers

The rotational inertia of a thin rod about one end is 1/3 ML2. What is the rotational inertia of the same rod about a point loca

zlopas [31]

Answer:

The value is $I = 0.0932 ML ^2$

Explanation:

From the question we are told that

The rotational inertia about one end is $I_R = \frac{1}{3} ML^2$

The location of the axis of rotation considered is $d = 0.4 L$

Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is $0.4 M$

Generally the length of the rod from the its beginning to the axis of rotation consider is

$k = 1 - 0.4 L = 0.6L$

Generally the mass of the portion of the rod from the its beginning to the axis of rotation consider is

$m = 1- 0.4 M = 0.6 M$

Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is

$I_{R1} = \frac{1}{3} (0.6 M )(0.6L)^2$

$I_{R1} = \frac{1}{3} (0.6 M )L^2 0.6^2$

Generally the rotational inertia about the axis of rotation consider for the second portion of the rod is

$I_{R2} = \frac{1}{3} (0.6 M )(0.6L)^2$

=> $I_{R2} = \frac{1}{3} (0.4 M )(0.4L)^2$

=> $I_{R2} = \frac{1}{3} (0.4 M )L^2 0.4^2$

Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is

$I = \frac{1}{3} (0.6 M )L^2 0.6^2 + \frac{1}{3} (0.4 M )L^2 0.4^2$

=> $I = \frac{1}{3} ML ^2 [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ]$

=> $I = 0.0932 ML ^2$

8 0

3 years ago

In which of the following cases is work being done on an object? Question 2 options: Pushing against a locked door Carrying a bo

Serggg [28]

As we know that work done is defined as product of force and displacement in the direction of applied force

so we can say it is

$W = F.d$

or we can say

$W = Fdcos\theta$

so we will have following options

A)Pushing against a locked door

There is no work done in above case as there will be no displacement in that case so work done is ZERO

B) Carrying a box down a corridor

While we move done and carrying a box then work done by gravity on the box while our work done is negative work.

C) Pulling a trailer up a hill

In above case our work done to pull a box up a hill is positive work as we need to work against gravity

D) Suspending a heavy weight with a strong chain

During suspension there is no displacement so there will be no work done in above case

so correct answer will be

<em>Pulling a trailer up a hill</em>

6 0

4 years ago

Read 2 more answers

A sinusoidal wave has the following wave function: y(x,t) = (2.5 m) sin[(3.0 m-1) x – (24 s-1) t + π/2] What is the frequency of

Vera_Pavlovna [14]

Answer:

3.82 Hertz

Explanation:

$y = 2.5 Sin\left (3x-24t+\frac{\pi }{2} \right )$

This is the equation of a wave which varies sinusoidally.

The standard equation of a wave is given by

$y = A Sin\left ( kx-\omega t+\phi \right )$

where, A be the amplitude of the wave, k be the wave number, x be the displacement of wave, ω be the angular frequency and t be the time taken, and Ф be the phase angle.

now compare the given equation by the standard equation, we get

k = 3

ω = 24

Ф = π / 2

So, the angular frequency = 24

The relation between the angular frequency and the frequency is given by

ω = 2 x π x f

24 = 2 x 3.14 x f

f = 3.82 Hertz

4 0

3 years ago