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3 years ago

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The radial component of acceleration of a particle moving in a circular path is always:________ a. negative. b. directed towards

the center of the path. c. perpendicular to the transverse component of acceleration d. all of the above

1 answer:

lesya [120]3 years ago

7 0

Answer:

d. all of the above

Explanation:

There are two components of acceleration for a particle moving in a circular path, radial and tangential acceleration.

The radial acceleration is given by;

$a_r = \frac{V^2}{R}$

Where;

V is the velocity of the particle

R is the radius of the circular path

This radial acceleration is always directed towards the center of the path, perpendicular to the tangential acceleration and negative.

Therefore, from the given options in the question, all the options are correct.

d. all of the above

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A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic

DiKsa [7]

Answer:

The answer is below

Explanation:

1) The synchronous speed of an induction motor is the speed of the magnetic field of the stator. It is given by:

$n_s=\frac{120f_s}{p}\\ Where\ p\ is \ the \ number\ of\ machine\ pole, f_s\ is\ the\ supply \ frequency\\and\ n_s\ is \ the \ synchronous\ speed(speed \ of\ stator\ magnetic \ field)\\Given: f_s=60\ Hz, p=4. Therefore\\\\n_s=\frac{120*60}{4}=1800\ rpm$

2) The speed of the rotor is the motor speed. The slip is given by:

$Slip=\frac{n_s-n_m}{n_s}. \\ n_m\ is\ the \ motor\ speed(rotor\ speed)\\Slip = 0.05, n_s= 1800\ rpm\\ \\0.05=\frac{1800-n_m}{1800}\\\\ 1800-n_m=90\\\\n_m=1800-90=1710\ rpm$

3) The frequency of the rotor is given as:

$f_r=slip*f_s\\f_r=0.04*60=2.4\ Hz$

4) At standstill, the speed of the motor is 0, therefore the slip is 1.

The frequency of the rotor is given as:

$f_r=slip*f_s\\f_r=1*60=60\ Hz$

6 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.

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3 years ago

You installed a new 40 gallon water heater with a 54,000 BTUh burner. The underground water temperature coming into the house is

kipiarov [429]

14256000. Kanjiuijhgg

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3 years ago

Read 2 more answers

The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet

alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

$T_{h_i}=200^{\circ}C$

$T_{h_o}=120^{\circ}C$

$T_{c_i}=100^{\circ}C$

$T_{c_o}=125^{\circ}C$

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

$m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]$

$\frac{m_hc_{ph}}{m_cc_{pc}}$ = $\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )$

we can see that heat capacity of hot fluid is minimum

Also from energy balance

$Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )$

$NTU=\frac{UA}{\left ( mc_p\right )_{h}}$ = $\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}$

$T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}$

$T_m=41.63^{\circ}C$

NTU=1.921

$And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}$

$\epsilon =\frac{1-0.2669}{1-0.0834}$

$\epsilon =0.799$

5 0

3 years ago

A cone penetration test was carried out in normally consolidated sand, for which the results are summarized below: Depth (m) Con

Cerrena [4.2K]

Answer:

hello your question is incomplete attached below is the missing equation related to the question

answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°

Explanation:

<u>Determine the friction angle at each depth</u>

attached below is the detailed solution

To calculate the vertical stress = depth * unit weight of sand

also inverse of Tan = Tan^-1

also qc is in Mpa while σ0 is in kPa

Friction angle at each depth

2 meters = 40.389°

3.5 meters = 38.987°

5 meters = 38.022°

6.5 meters = 39.869°

8 meters = 40.265°

6 0

3 years ago