How are airplanes artificially pressurized?

A leaky 10-kg bucket is lifted from the ground to a height of 10 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

sladkih [1.3K]

Answer:

The work done shall be 14715 Joules

Explanation:

The work done by a force 'F' in a displacement 'dy' is given by

$W=m(y)g\times dy$

At any position 'y' the weight shall be sum of weft of water and weight of string

$\therefore m(y)=m_{water}(y)+m_{string}(y)\\\\m(y)=30(1-\frac{y}{10})+0.9y$

Thus applying values we get

$W=\int m(y)g\times dy\\\\W=\int_{0}^{10}(30(1-\frac{y}{10}+0.9y)\times g)dy\\\\Solving\\W=294.3\int_{0}^{10}(1-\frac{y}{10}+0.9y)dy\\\\W=14715J$

8 0

3 years ago

Under the assumption that the beam is a rectangular cantilever beam that is free to vibrate, the theoretical first natural frequ

BartSMP [9]

Answer:

a) Δf = 0.7 n , e) f = (15.1 ± 0.7) 10³ Hz

Explanation:

This is an error about the uncertainty or error in the calculated quantities.

Let's work all the magnitudes is the SI system

The frequency of oscillation is

f = n / 2π L² √( E /ρ)

where n is an integer

Let's calculate the magnitude of the oscillation

f = n / 2π (0.2335)² √ (210 10⁹/7800)

f = n /0.34257 √ (26.923 10⁶)

f = n /0.34257 5.1887 10³

f = 15.1464 10³ n

a) We are asked for the uncertainty of the frequency (Df)

Δf = | df / dL | ΔL + df /dE ΔE + df /dρ Δρ

in this case no error is indicated in Young's modulus and density, so we will consider them exact

ΔE = Δρ = 0

Δf = df /dL ΔL

df = n / 2π √E /ρ | -2 / L³ | ΔL

df = n / 2π 5.1887 10³ | 2 / 0.2335³) 0.005 10⁻³

df = n 0.649

Absolute deviations must be given with a single significant figure

Δf = 0.7 n

b, c) The uncertainty with the width and thickness of the canteliver is associated with the density

In your expression there is no specific dependency so the uncertainty should be zero

The exact equation for the natural nodes is

f = n / 2π L² √ (E e /ρA)

where A is the area of the cantilever and its thickness,

In this case, they must perform the derivatives, calculate and approximate a significant figure

Δf = | df / dL | ΔL + df /de Δe + df /dA ΔA

Δf = 0.7 n + n 2π L² √(E/ρ A) | ½ 1/√e | Δe

+ n / 2π L² √(Ee /ρ) | 3/2 1√A23 |

the area is

A = b h

A = 24.9 3.3 10⁻⁶

A = 82.17 10⁻⁶ m²

DA = dA /db ΔB + dA /dh Δh

dA = h Δb + b Δh

dA = 3.3 10⁻³ 0.005 10⁻³ + 24.9 10⁻³ 0.005 10⁻³

dA = (3.3 + 24.9) 0.005 10⁻⁶

dA = 1.4 10⁻⁷ m²

let's calculate each term

A ’= n / 2π L² √a (E/ρ A) | ½ 1 /√ e | Δe

A ’= n/ 2π L² √ (E /ρ) | ½ 1 / (√e/√ A) |Δe

A ’= 15.1464 10³ n ½ 1 / [√ (24.9 10⁻³)/ √ (81.17 10⁻⁶)] 0.005 10⁻³

A '= 0.0266 n

A ’= 2.66 10⁻² n

A ’’ = n / 2π L² √ (E e /ρ) | 3/2 1 /√A³ |

A ’’ = n / 2π L² √(E /ρ) √ e | 3/2 1 /√ A³ | ΔA

A ’’ = n 15.1464 10³ 3/2 √ (24.9 10⁻³) /√ (82.17 10⁻⁶) 3 1.4 10⁻⁷

A ’’ = n 15.1464 1.5 1.5779 / 744.85 1.4 10⁴

A ’’ = 6,738 10²

we write the equation of uncertainty

Δf = n (0.649 + 2.66 10⁻² + 6.738 10²)

The uncertainty due to thickness is

Δf = 3 10⁻² n

The uncertainty regarding the area, note that this magnitude should be measured with much greater precision, specifically the height since the errors of the width are very small

Δf = 7 10² n

d) Δf = 7 10² n

e) the natural frequency n = 1

f = (15.1 ± 0.7) 10³ Hz

7 0

3 years ago

Helpppppppp please !!!!

agasfer [191]

Answer:

second one is correct that is right

7 0

3 years ago

Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 21.0 cm and carries a

Burka [1]

Solution :

a). B at the center :

$=\frac{u\times I}{2R}$

Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.

Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE

b). Also, the sum of the fields must be zero.

Therefore,

$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$

So,

$\frac{I_1}{d_1}= \frac{I_2}{d_2}$

$=\frac{16}{21}=\frac{I_2}{32}$

$I_2=24.38$ A

Therefore, the current in the outer wire is 24.38 ampere.

3 0

3 years ago

Read 2 more answers

A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most

Oxana [17]

Answer:

A) $I_{total}$ = 1.44 kg m², B) moment of inertia must increase

Explanation:

The moment of inertia is defined by

I = ∫ r² dm

For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is

I = ½ m R²

A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is

I = $I_{cm}$ + m D²

Let's apply these equations to our case

The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms

$I_{total}$ = $I_{body}$ + 2 $I_{arm}$

$I_{body}$ = ½ M R²

The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body

M = 7/8 m total

M = 7/8 64

M = 56 kg

The mass of the arms is

m’= 1/8 m total

m’= 1/8 64

m’= 8 kg

As it has two arms the mass of each arm is half

m = ½ m ’

m = 4 kg

The arms are very thin, we will approximate them as a particle

$I_{arm}$ = M D²

Let's write the equation

$I_{total}$ = ½ M R² + 2 (m D²)

Let's calculate

$I_{total}$ = ½ 56 0.20² + 2 4 0.20²

$I_{total}$ = 1.12 + 0.32

$I_{total}$ = 1.44 kg m²

b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase

6 0

3 years ago