How are weather and climate different

Answer:

Your mobile phone has more computing power than the computers used for the Apollo 11 moon landing.

Mobile phones have to “work harder” to get a signal if you are in a moving vehicle.

The first mobile phone was made in 1973.

The first mobile phones that went on sale in 1983 cost nearly $4,000 each.

In 2012 Apple sold 340,000 phones per day.

4 out of 10 Brits admit to snooping on their partners phone.

Out of the 53% of snoopers that found incriminating evidence on their partner’s phone, 5% went on to terminate their relationship.

Waterproof mobile phones came to market because Japanese youngsters like to use them in the shower.

Apparently mobile phones have 18 times more bacteria on them than toilet handles!

Phubbing describes the act of snubbing someone by using your mobile phone in their company.

In 2015 more people died from taking selfies than shark attacks.

Teenagers that use a phone more than 2 hours a day increase their risk of depression and anxiety.

Nomobophobia is severe anxiety caused by the thought or act of losing your phone or running out of battery.

Explanation:

Answer:

T = 3460 K

Explanation:

See attachment for calculation.

Since the temperature we have is above the melting point of the metal, then we can conclude that it is too high for the diffusion process to be possible.

Answer:

Resolution of Fourier transform = 2Hz

Explanation:

Given:

Sample signals = 16,384 samples/second

Number of data points = 8,192 oints

Find:

Resolution of Fourier transform

Computation:

Resolution of Fourier transform = fs / N

Resolution of Fourier transform = 16,384 / 8,192

Resolution of Fourier transform = 2Hz

Answer:

116,090 Btus

Explanation:

Answer:

Speed V1 at the beginning of the skid  = 57.075 mph

Explanation:

Calculate the V1 speed at the beginning of the skid

V1 ( speed before brake was applied ) = ?

distance travelled by Vehicle 1 = 195 ft

curve tangent for vehicle 1 =  - 3% = - 0.03

coefficient of friction = 0.48

speed at Impact = 25 mph

To determine the V1 speed at the beginning of the skid we have to apply AASTHO

d1 = [tex]( \frac{VA^{2} - VA^{12} }{30(F+N)} )[/tex]

195 =  [tex](\frac{VA^2- 25^2}{30(0.48-0.03)} )[/tex]

Hence : VA = 57.075 mph

Answer:

Mach number = 0.68168

The flow should be treated as compressible.

Explanation:

Given that:

The altitude of a commercial jet = 35000

The properties of air at that given altitude are as follows:

Pressure = 24.577 kPa

Temperature T = 50.78176° C

Temperature T = ( 50.78176 + 273 )K = 328.78176 K

[tex]\varphi = 0.38428 \ kg/m^3[/tex]

The velocity is also given as: 550 mph = 245.872 m/s

Therefore, the sonic velocity is firstly determined by using the formula:

[tex]a = \sqrt{ \vartheta \times R \times T\[/tex]

[tex]a = \sqrt{1.4 \times 287 \times 323.78176[/tex]

[tex]a = \sqrt{130095.5112[/tex]

a = 360.68755 m/s

Then, we can calculate the Mach number by using the expression:

[tex]{Mach \ number = \dfrac{V}{a}}[/tex]

[tex]Mach \ number = \dfrac{245.872}{360.68755}[/tex]

Mach number = 0.68168

b) Ideally, all flows are compressible because the Mach number is greater than 0.3, suppose the Mach number is lesser than 0.3, then it is incompressible.

Answer:

The moment of inertia of the rotor is approximately [tex]1.105\times 10^{-6}[/tex] kilogram-square meters.

The rotor inertia may differ from these assumption due to differences in the shape of cross section.

Explanation:

We assume that rotor can be represented as a solid cylinder of radius [tex]r[/tex], length [tex]l[/tex], made of cooper ([tex]\rho = 8960\,\frac{kg}{m^{3}}[/tex]) and whose axis of rotation passes through its center of mass and is parallel to its cross section. By definition of Moment of Inertia and Theorem of Parallel Axes, the moment of inertia of the rotot is:

[tex]I = \frac{1}{4}\cdot \rho \cdot \left(\frac{\pi}{4} \right) \cdot R^{3}\cdot (3\cdot R^{2}+L^{2})[/tex]

[tex]I = \frac{\pi}{16}\cdot \rho \cdot R^{3}\cdot (3\cdot R^{2}+L^{2})[/tex] (Eq. 1)

Where:

[tex]\rho[/tex] - Density of copper, measured in kilograms per cubic meter.

[tex]R[/tex] - Radius of the rotor, measured in meters.

[tex]L[/tex] - Length of the rotor, measured in meters.

[tex]I[/tex] - Moment of inertia, measured in kilogram-square meters.

If we know that [tex]\rho = 8960\,\frac{kg}{m^{3}}[/tex], [tex]L = 25\times 10^{-3}\,m[/tex] and [tex]R = 8.98\times 10^{-3}\,m[/tex], the estimated moment of inertia of the rotor is:

[tex]I = \frac{\pi}{16}\cdot \left(8960\,\frac{kg}{m^{3}} \right)\cdot (8.98\times 10^{-3}\,m)^{3}\cdot [3\cdot (8.98\times 10^{-3}\,m)^{2}+(25\times 10^{-3}\,m)^{2}][/tex]

[tex]I \approx 1.105\times 10^{-6}\,kg\cdot m^{2}[/tex]

The moment of inertia of the rotor is approximately [tex]1.105\times 10^{-6}[/tex] kilogram-square meters.

From D'Alembert's Formula we know that net force of rigid bodies experimenting rotation equals the product of moment of inertia and angular acceleration. In this case, the purpose is minimizing moment of inertia and it is done by modifying the shape of the cross section so that rotor could be aerodynamically more efficient.

Answer:

Option B: An MSDS

Explanation:

A dictionary is used to check up the meaning of general words and not for checking if a substance being used is hazardous. Option A is wrong.

MSDS means "Material Safety Data Sheet" and it contains documents with information that relates to occupational health & safety for checking various substances and products. Thus, option B is correct.

SAE stands for Society of Automotive Engineering and their standards pertain to mainly Automobiles. Thus option C is wrong.

EPA guidelines are mainly for checking facility and environmental health and safety compliance. Thus, option D is wrong.

Answer:

carried courage buried

Explanation:

its Filipino

Answer:

a Network Engineer (architect)

Explanation:

Indeed, as a Network Engineer, we would expect Nicole to be in charge of planning, setting up, and managing the software and hardware components of the computer networks so that they function as intended.

A Network Engineer therefore would be responsible for setting up the networking capabilities and policies that will govern the workstations when connected to the company network.

Answer:

[tex]7304g/cm^3[/tex]

Explanation:

Density is the mass per unit volume, it is usually measured in kg/m³.

The density of a crystal structure is given by:

[tex]\rho=\frac{mass\ of\ atoms\ in\ unit\ cell}{volume\ unit\ cell} =\frac{nA}{V_cN_A} \\\\\rho=density, n= number\ of\ atoms/unit=1\ atom/unit\ cell,A=atomic\ weight=70.4\ g/mol,N_A=avogadro\ constant=6.023*10^{23}atoms/mol,V=volume=2R^3=[2(0.126*10^{-8})]^3cm\\\\\rho=\frac{1*70.4}{[2(0.126*10^{-8})]^3*6.023*10^{23}} =7304g/cm^3[/tex]

Answer:

7.307g/cm^3

Explanation:

The other guy is mostly right but his conversion factor was off. It should be 1.26x10^-8 on the bottom

Answer:

The model system will need water flowing at a velocity of 2.07 meters per second to guarantee kinematic similarity in the form of equal Reynolds numbers.

Explanation:

The Reynolds number ([tex]Re_{D}[/tex]) is a dimensionless criterion use for flow regime of fluids, which is defined as:

[tex]Re_{D} = \frac{\rho \cdot v\cdot D}{\mu}[/tex] (Eq. 1)

Where:

[tex]\rho[/tex] - Density, measured in kilograms per cubic meter.

[tex]\mu[/tex] - Dynamic viscosity, measured in kilograms per meter-second.

[tex]v[/tex] - Average flow velocity, measured in meters per second.

[tex]D[/tex] - Pipe diameter, measured in meters.

We need to find the equivalent velocity of water used in the prototype system. In this case, we assume that [tex]Re_{D,gas} = Re_{D,w}[/tex]. That is:

[tex]\frac{\rho_{w}\cdot v_{w}\cdot D_{w}}{\mu_{w}} = \frac{\rho_{gas}\cdot v_{gas}\cdot D_{gas}}{\mu_{gas}}[/tex] (Eq. 2)

Where subindex [tex]w[/tex] is used for water and [tex]gas[/tex] for gasoline.

If we know that [tex]\rho_{gas} = 690\,\frac{kg}{m^{2}}[/tex], [tex]\mu_{gas} = 0.006\,\frac{kg}{m\cdot s}[/tex], [tex]v_{gas} = 0.5\,\frac{m}{s}[/tex], [tex]D_{gas} = 1\,m[/tex], [tex]\rho_{w} = 1000\,\frac{kg}{m^{3}}[/tex], [tex]\mu_{w} = 0.0018\,\frac{kg}{m\cdot s}[/tex] and [tex]D_{w} = 0.05\,m[/tex], then we get the following formula:

[tex]57500 = 27777.778\cdot v_{w}[/tex]

The fluid velocity for the prototype system is:

[tex]v_{w} = 2.07\,\frac{m}{y}[/tex]

The model system will need water flowing at a velocity of 2.07 meters per second to guarantee kinematic similarity in the form of equal Reynolds numbers.

Answer:

The estimated batching production is 210 cy/hr

Explanation:

Given;

capacity of the batching chamber of the plant, Q = 12.5 cy

average batching cycle time, t = 3 min

plant efficiency, n = 84 %

The estimated batching production is calculated as efficiency of the machine multiplied by the production rate in hour. This is given by the solution below;

Batching production = (nQ )/ t

[tex]Batching \ production = 0.84(\frac{12.5 \ cy}{3 \ min} *\frac{60 \ min}{1 \ hr})\\\\ Batching \ production = 210 \ cy/hr[/tex]

Therefore, the estimated batching production is 210 cy/hr

Answer:

Explanation:

7. False

8. True

9. True

10.True

11. True

Answer:

It's equalization of the system

Explanation: If there was a leak it would likely leak all the time even if the car was turned off. Plus, a system leak bad enough to hear would drain the system of refrigerant very quickly and would no longer cool.

Answer:

72 MPa

Explanation:

Critical resolved shear stress = 36 MPa

calculate the maximum possible yield strength for a single crystal of the metal

first we have to express the critical resolved shear stress as

бy = 2т[tex]_{critical}[/tex]

given that the minimum stress needed to introduce yielding curve will occur  at Ψ = λ = 45

where : бy = yielding stress

             t[tex]_{critical}[/tex] = 36MPa

therefore the maximum possible yield strength

= (2) * (36)

= 72 MPa

Answer:

a) According to the maximum-shear-stress failure theory, the static factor of safety of the shaft is 2.440.

b) According to the distortion-energy failure theory, the static factor of safety of the shaft is 2.816.

Explanation:

First, we need to determine the torque experimented by the shaft ([tex]T[/tex]), measured in kilonewton-meters, whose formula is described:

[tex]T = \frac{\dot W}{\omega}[/tex] (Eq. 1)

Where:

[tex]\dot W[/tex] - Power, measured in kilowatts.

[tex]\omega[/tex] - Angular velocity, measured in radians per second.

If we know that [tex]\dot W = 10\,kW[/tex] and [tex]\omega = 20.944\,\frac{rad}{s}[/tex], then the torque experimented by the shaft:

[tex]T = \frac{10\,kW}{20.944\,\frac{rad}{s} }[/tex]

[tex]T =0.478\,kN\cdot m[/tex]

Let consider that shaft has a circular form, such that shear stress is determined by the following formula:

[tex]\tau = \frac{16\cdot T}{\pi\cdot D^{3}}[/tex] (Eq. 2)

Where:

[tex]D[/tex] - Diameter of the shaft, measured in meters.

[tex]\tau[/tex] - Torsional shear stress, measured in kilopascals.

If we know that [tex]D = 0.03\,m[/tex] and [tex]T =0.478\,kN\cdot m[/tex], the torsional shear stress is:

[tex]\tau = \frac{16\cdot (0.478\,kN\cdot m)}{\pi\cdot (0.03\,m)^{3}}[/tex]

[tex]\tau \approx 90164.223\,kPa[/tex]

a) According to the maximum-shear-stress failure theory, we get that maximum shear stress limit is:

[tex]S_{ys} = 0.5\cdot S_{ut}[/tex] (Eq. 3)

Where:

[tex]S_{ys}[/tex] - Ultimate shear stress, measured in kilopascals.

[tex]S_{ut}[/tex] - Ultimate tensile stress, measured in kilopascals.

If we know that [tex]S_{ut} = 440\times 10^{3}\,kPa[/tex], the ultimate shear stress of the material is:

[tex]S_{ys} = 0.5\cdot (440\times 10^{3}\,kPa)[/tex]

[tex]S_{ys} = 220\times 10^{3}\,kPa[/tex]

Lastly, the static factor of safety of the shaft ([tex]n[/tex]), dimensionless, is:

[tex]n = \frac{S_{ys}}{\tau}[/tex] (Eq. 4)

If we know that [tex]S_{ys} = 220\times 10^{3}\,kPa[/tex] and [tex]\tau \approx 90164.223\,kPa[/tex], the static factor of safety of the shaft is:

[tex]n = \frac{220\times 10^{3}\,kPa}{90164.223\,kPa}[/tex]

[tex]n = 2.440[/tex]

According to the maximum-shear-stress failure theory, the static factor of safety of the shaft is 2.440.

b) According to the distortion-energy failure theory, we get that maximum shear stress limit is:

[tex]S_{ys} = 0.577\cdot S_{ut}[/tex] (Eq. 5)

If we know that [tex]S_{ut} = 440\times 10^{3}\,kPa[/tex], the ultimate shear stress of the material is:

[tex]S_{ys} = 0.577\cdot (440\times 10^{3}\,kPa)[/tex]

[tex]S_{ys} = 253.88\times 10^{3}\,kPa[/tex]

Lastly, the static factor of safety of the shaft is:

[tex]n = \frac{253.88\times 10^{3}\,kPa}{90164.223\,kPa}[/tex]

[tex]n = 2.816[/tex]

According to the distortion-energy failure theory, the static factor of safety of the shaft is 2.816.

Answer:

a) 4 hectómetros cuadrados equivalen a 400 decámetros cuadrados.

b) 21345 centímetros cuadrados equivalen a 2,135 metros cuadrados.

c) 0,592 kilómetros cuadrados equivalen a 592000 metros cuadrados.

d) 0,102 metros cuadrados equivalen a 1020 centímetros cuadrados.  

e) 23911 kilómetros cuadrados equivalen 2391100 hectómetros cuadrados.

Explanation:

a) 4 hectómetros cuadrados a decámetros cuadrados:

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un hectómetro cuadrado equivale a 100 decámetros cuadradps. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

[tex]x = 4\,Hm^{2}\times\frac{100\,Dm^{2}}{1\,Hm^{2}}[/tex]

[tex]x = 400\,Dm^{2}[/tex]

4 hectómetros cuadrados equivalen a 400 decámetros cuadrados.

b) 21345 centímetros cuadrados a metros cuadrados:

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un metro cuadrado equivale a 10000 centímetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

[tex]x = 21345\,cm^{2}\times \frac{1\,m^{2}}{10000\,cm^{2}}[/tex]

[tex]x = 2,135\,m^{2}[/tex]

21345 centímetros cuadrados equivalen a 2,135 metros cuadrados.

c) 0,592 kilómetros cuadrados a metros cuadrados:

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un kilómetro cuadrado equivale a 1000000 metros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

[tex]x = 0,592\,km^{2}\times \frac{1000000\,m^{2}}{1\,km^{2}}[/tex]

[tex]x = 592000\,m^{2}[/tex]

0,592 kilómetros cuadrados equivalen a 592000 metros cuadrados.

d) 0,102 metros cuadrados a centímetros cuadrados:

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un metro cuadrado equivale a 10000 centímetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

[tex]x = 0,102\,m^{2}\times \frac{10000\,cm^{2}}{1\,m^{2}}[/tex]

[tex]x = 1020\,cm^{2}[/tex]

0,102 metros cuadrados equivalen a 1020 centímetros cuadrados.

e) 23911 kilómetros cuadrados a hectómetros cuadrados:

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un kilómetro cuadrado equivale a 100 hectómetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

[tex]x = 23911\,km^{2}\times \frac{100\,Hm^{2}}{1\,km^{2}}[/tex]

[tex]x = 2391100\,Hm^{2}[/tex]

23911 kilómetros cuadrados equivalen 2391100 hectómetros cuadrados.

Answer:

  no

Explanation:

No. More power is dissipated when the transistor is in its active region. In general, transistors in switching circuits are biased either "on" or "off". Time spent in the active region is minimized.

_____

On the other hand, speed can be enhanced if the transistors are active. So, it's a speed/power trade-off. Usually power is of more interest, particularly when there are millions of switching circuits. However, in certain applications, speed may be the priority, so the transistor will be biased in its active region.