Which of the following is critical when performing maintenance?

A 11-cm-diameter horizontal jet of water with a velocity of 40 m/s relative to the ground strikes a flat plate that is moving in

Reika [66]

Answer:

F = 8552.7N

Explanation:

We need first our values, that are,

$V_{jet} = 40m/s\\V_{Plate} = 10m/s \\D = 11cm$

We start to calculate the relative velocity, that is,

$V_r = V_{jet}-V_{plate}\\V_r = (40)-(10)\\V_r = 30m/s$

With the relative velocity we can calculate the mass flow rate, given by,

$\dot{m}_r = \rho A V_r$

$\dot{m}_r = (1000)(30) \frac{\pi (0.11)^2}{4}$

$\dot{m}_r = 285.09kg/s$

We need to define the Force in the direction of the flow,

$\sum\vec{F} = \sum_{out} \beta\dot{m}\vec{V} - \sum_{in} \beta\dot{m} \vec{V}\\$

$F = \dot{m}V_r$

$F = (285.09Kg/s)(30)$

$F = 8552.7N$

8 0

2 years ago

You are using a Jupyter Notebook to explore data in a DataFrame named productDF. You want to write some inline SQL by using the

defon

You need to explain it more simple as everyone is clueless

7 0

2 years ago

Vẽ thủ tục cho một cuộc gọi thuê bao

shusha [124]

Lo siento, no sé qué estás diciendo.

8 0

2 years ago

There is an electric field near the Earth's surface whose magnitude is about 145 V/m . How much energy is stored per cubic meter

weqwewe [10]

Answer:

u_e = 9.3 * 10^-8 J / m^3 ( 2 sig. fig)

Explanation:

Given:

- Electric Field strength near earth's surface E = 145 V / m

- permittivity of free space (electric constant) e_o = 8.854 *10^-12 s^4 A^2 / m^3 kg

Find:

- How much energy is stored per cubic meter in this field?

Solution:

- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:

u_e = 0.5*e_o * E^2

- Plug in the values given:

u_e = 0.5*8.854 *10^-12 *145^2

u_e = 9.30777 * 10^-8 J/m^3

5 0

3 years ago

A well-insulated, rigid tank has a volume of 1 m3and is initially evacuated. A valve is opened,and the surrounding air enters at

DiKsa [7]

Answer:

0.5 kW

Explanation:

The given parameters are;

Volume of tank = 1 m³

Pressure of air entering tank = 1 bar

Temperature of air = 27°C = 300.15 K

Temperature after heating = 477 °C = 750.15 K

V₂ = 1 m³

P₁V₁/T₁ = P₂V₂/T₂

P₁ = P₂

V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³

$dQ = m \times c_p \times (T_2 -T_1)$

For ideal gas, $c_p$ = 5/2×R = 5/2*0.287 = 0.7175 kJ

PV = NKT

N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)

N = 9.66×10²⁴

Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles

The average mass of one mole of air = 28.8 g

Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg

∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ

The power input required = The rate of heat transfer = 149.211/(60*5)

The power input required = 0.49737 kW ≈ 0.5 kW.

3 0

2 years ago