All categories

2 years ago

How to tie shoe please

1 answer:

4 0

Step 1: Unknot Shoelaces. Make sure that the two ends of the shoelace are completely untangled and free of knots past the tongue. ...
Step 2: Create Overhand Knot. ...
Step 3: Finish the Basic Shoe Knot. ...
Step 4: Create Double Knot. ...
Step 5: Finished!

You might be interested in

A sheet of steel 4.4 mm thick has nitrogen atmospheres on both sides at 1200°C and is permitted to achieve a steady-state diffus

VARVARA [1.3K]

Answer:

0.544×10–³

Explanation:

Please see the attached file for the solution

7 0

3 years ago

A food department is kept at -12 °C by a refrigerator in an environment at 30 °C. The total heat gain to the food department is

Nataly [62]

Answer: P = 0.416 kW

Explanation:

taken a step by step process to solving this problem.

we have that from the question;

the amount of heat rejected Qn = 4800 kJ/h

the cooling effect is Ql = 3300 kJ/h

Applying the first law of thermodynamics for this system gives us

Шnet = Qn -Ql

Шnet = 4800 - 3300 = 1500 kJ/h

Next we would calculate the coefficient of performance of the refrigerator;

COPr = Desired Effect / work output = Ql / Шnet = 3300/1500 = 2.2

COPr = 2.2

The Power as required gives;

P = Qn - Ql = 4800 - 3300 = 1500 kJ/h = 0.416

P = 0.416 kW

cheers i hope this helps!!!!1

5 0

3 years ago

A pressure cylinder has an outer diameter 200 mm, maximum external pressure 4 MPa, and maximum allowable shear stress 27.5 MPa.

ludmilkaskok [199]

Answer:

The minimum value of wall thickness t=3.63 mm.

Explanation:

Given:

D=200 mm

P=4 MPa

t= Wall thickness

maximum shear stress=27.5 MPa

We know that

hoop stress $\sigma _{h}=\frac{Pd}{2t}$

Longitudinal stress $\sigma _{l}=\frac{Pd}{4t}$

So maximum shear tress in plane $\tau _{max}=\dfrac{\sigma _h-\sigma _l}{2}$

$\tau _{max}=\dfrac{Pd}{8t}$

Now by putting the value

$27.5=\dfrac{4\times 200}{8t}$

So t=3.36 mm

The minimum value of wall thickness t=3.63 mm.

4 0

3 years ago

Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−

sergeinik [125]

Answer:

Power output, $P_{out} = 178.56 kW$

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, $T = 350^{\circ}C$

Diameter of pipe, d = 8 cm = 0.08 m

Mass flow rate, $\dot{m} = 0.1 kg.s^{- 1}$

Diameter of exhaust pipe, $d_{h} = 15 cm = 0.15 m$

Pressure at exhaust, P' = 50 kPa

temperature, T' = $100^{\circ}C$

Solution:

Now, calculation of the velocity of fluid at state 1 inlet:

$\dot{m} = \frac{Av_{i}}{V_{1}}$

$0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}$

$0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}$

$v_{i} = 3.986 m/s$

Now, eqn for compressible fluid:

$\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}$

Now,

$\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}$

$\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}$

$\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}$

$v_{e} = 19.33 m/s$

Now, the power output can be calculated from the energy balance eqn:

$P_{out} = -\dot{m}W_{s}$

$P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}$

$P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW$

4 0

3 years ago

ANSWER FAST PLEASE!!! WILL MARK BRAINLIEST!!!!!!

Phoenix [80]

Answer:

Phuong works on a research project and creates a report for her boss.

5 0

3 years ago

Read 2 more answers