A body of weight 300N is lying rough
horizontal plane having
a Coefficient of friction as 0.3
Find the magnitude of the forces which can move the
body while acting at an angle of 25 with the horizonted​

Answer:

i) Primary sensing element.

ii) Variable conversion element.

iii) Variable manipulation element.

Explanation:

Answer:

The main functional elements of a measurement system are:

i) Primary sensing element.

ii) Variable conversion element.

iii) Variable manipulation element.

v) Data transmission element.

Answer:

L = Henry

C = Farad

Explanation:

The electrical parameter represented as L is the inductance whose unit is Henry(H).

The electrical parameter represented as C is the inductance whose unit is Farad

Resonance frequency occurs when the applied period force is equal to the natural frequency of the system upon which the force acts :

To obtain :

At resonance, Inductive reactance = capacitive reactance

Equate the inductive and capacitive reactance

Inductive reactance(Xl) = 2πFL

Capacitive Reactance(Xc) = 1/2πFC

Inductive reactance(Xl) = Capacitive Reactance(Xc)

2πFL = 1/2πFC

Multiplying both sides by F

F * 2πFL = F * 1/2πFC

2πF²L = 1/2πC

Isolating F²

F² = 1/2πC2πL

F² = 1/4π²LC

Take the square root of both sides to make F the subject

F = √1 / √4π²LC

F = 1 /2π√LC

Hence, the proof.

Answer:

2901 vehicles

Explanation:

We are given;

Percentage of large trucks & buses; p_t = 7% = 0.07

Percentage of recreational vehicles; p_r = 3% = 0.03

PHF = 0.90

Driver population adjustment; f_p = 0.92

First of all, let's Calculate the heavy vehicle factor from the formula;

f_hv = 1/[1 + p_t(e_t - 1) + p_r(e_r - 1)]

Where;

e_t = passenger car equivalents for trucks and buses

e_r = passenger car equivalents for recreational vehicles

From the table attached, for a mountainous terrain, e_t = 6 and e_r = 4. Thus;

f_hv = 1/[1 + 0.07(6 - 1) + 0.03(4 - 1)]

f_hv = 1.44

Let's now calculate the initial hourly volume from the formula;

v_p = V1/(PHF × N × f_hv × f_p)

Where;

v_p = 15-minute passenger-car equivalent flow rate

V1 = hourly volume

N = number of lanes in each direction

From online tables of LOS criteria for multilane freeway segments, v_p = 1300 pc/hr/ln

Thus;

1300 = V1/(0.9 × 3 × 1.44 × 0.92)

V1 = 1300 × (0.9 × 3 × 1.44 × 0.92)

V1 = 4650 veh/hr

Now, let's Calculate the final hourly volume;

From online sources, the maximum capacity of a 6 lane highway with free-flow speed of 50 mi/h is 2000 pc/hr/ln.

We are told the online peak-hour factor increases to 0.95 and so PHF = 0.95.

Thus;

2000 = V2/(0.95 × 3 × 1.44 × 0.92)

V2 = 2000(0.95 × 3 × 1.44 × 0.92)

V2 = 7551 veh/hr

Number of vehicles added to the highway = V2 - V1 = 7551 - 4650 = 2901 vehicles

Answer:

1. i. 20 Nm ii. 4.85 HP

2. 16.5 %

Explanation:

1) Determine the torque and the power developed at 4% slip when a reduced voltage of 340V is applied.

i. Torque

Since slip is constant at 4 %,torque, T ∝ V² where V = voltage

Now, T₂/T₁ = V₂²/V₁² where T₁ = torque at 690 V = P/2πN where P = power = 20 HP = 20 × 746 W = 14920 W, N = rotor speed = N'(1 - s) where s = slip = 4% = 0.04 and N' = synchronous speed = 120f/p where f = frequency = 60 Hz and p = number of poles = 4.

So, N' = 120 × 60/4 = 30 × 60 = 1800 rpm

So, N = N'(1 - s) = 1800 rpm(1 - 0.04) = 1800 rpm(0.96) = 1728 rpm = 1728/60 = 28.8 rps

So, T = P/2πN = 14920 W/(2π × 28.8rps) = 14920 W/180.96 = 82.45 Nm

T₂ = torque at 340 V, V₁ = 690 V and V₂ = 340 V

So, T₂/T₁ = V₂²/V₁²

T₂ = (V₂²/V₁²)T₁

T₂ = (V₂/V₁)²T₁

T₂ = (340 V/690 V)²82.45 Nm

T₂ = (0.4928)²82.45 Nm

T₂ = (0.2428)82.45 Nm

T₂ = 20.02 Nm

T₂ ≅ 20 Nm

ii. Power

P = 2πT₂N'

= 2π × 20 Nm × 28.8 rps

= 1152π W

= 3619.11 W

converting to HP

= 3619.11 W/746 W

= 4.85 HP

2) What must be the new slip for the motor to develop the same torque when the reduced voltage is applied

Since torque T ∝ sV² where s = slip and V = voltage,

T₂/T₁ = s₂V₂²/s₁V₁²

where T₁ = torque at slip, s₁ = 4% and voltage V₁ = 690 V and T₂ = torque at slip, s₂ = unknown and voltage V₂ = 340 V

If the torque is the same, T₁ = T₂ ⇒ T₂T₁ = 1

So,

T₂/T₁ = s₂V₂²/s₁V₁²

1 = s₂V₂²/s₁V₁²

s₂V₂² = s₁V₁²

s₂ = s₁V₁²/V₂²

s₂ = s₁(V₁/V₂)²

substituting the values of the variables into the equation, we have

s₂ = s₁(V₁/V₂)²

s₂ = 4%(690/340)²

s₂ = 4%(2.0294)²

s₂ = 4%(4.119)

s₂ = 16.47 %

s₂ ≅ 16.5 %

The seismic response of nuclear power plant structures is often calculated using lumped parameter methods. A finite element model of the structure is coupled to the soil with a spring-dashpot system used to represent the interaction process. The parameters of the interaction model are based on analytic solutions to simple problems which are idealizations of the actual problems of interest. The objective of the work reported in this paper is to compare predicted responses using the standard lumped parameter models with experimental data. These comparisons are shown to be good for a fairly uniform soil system and for loadings that do not result in nonlinear interaction effects such as liftoff. 7 references, 7 figures.

Answer:

Hydrostatic force = 41168 N

Explanation:

Complete question

A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water  so that the top is 4 ft below the surface. If the base is in the surface of water, find the force against onr side of the plate. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)

Let "x" be the side length submerged in water.

Then

w(x)/base = (4+3-x)/altitude

w(x)/5 = (4+3-x)/3

w(x) = 5* (7-x)/3

Hydrostatic force = 62.5 integration of  x * 4 * (10-x)/3 with limits from 4 to 7

HF = integration of 40x - 4x^2/3

HF = 20x^2 - 4x^3/9 with limit 4 to 7

HF = (20*7^2 - 4*7^(3/9))- (20*4^2 - 4*4^(3/9))

HF = 658.69 N *62.5 = 41168 N

Answer: hello some parts of your question is missing attached below is the missing information

The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d  is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube

answer : Total surface area = 3/2 * area of old radiator

Explanation:

we will use this relation

K = [tex]\frac{Qd }{A* change in T }[/tex]

change in T =  ΔT  

therefore New Area  ( A ) = 3/2 * area of old radiator

Given that the thermal conductivity is the same in the new and old radiators

Answer:

C = $0.0032 per day

Explanation:

We are given;

Dimension of cell phone; 50 mm × 45 mm × 20 mm

Temperature of charger; T1 = 33°C = 306K

Emissivity; ε = 0.92

convection heat transfer coefficient; h = 4.5 W/m².K

Room air temperature; T∞ = 22°C = 295K

Wall temperature; T2 = 20°C = 293 K

Cost of electricity; C = $0.18/kW.h

Chargers are usually in the form of a cuboid, and thus, surface Area is;

A = (50 × 45) + 2(50 × 20) + 2(45 × 20)

A = 6050 mm²

A = 6.05 × 10^(-3) m²

Formula for total heat transfer rate is;

E_t = hA(T1 - T∞) + εσA((T1)⁴ - (T2)⁴)

Where σ is Stefan Boltzmann constant with a value of; σ = 5.67 × 10^(-8) W/m².K⁴

Thus;

E_t = 4.5 × 6.05 × 10^(-3) (306 - 295) + (0.92 × 6.05 × 10^(-3) × 5.67 × 10^(-8)(306^(4) - 293^(4)))

E_t = 0.7406 W = 0.7406 × 10^(-3) KW

Now, we know C = $0.18/kW.h

Thus daily cost which has 24 hours gives;

C = 0.18 × 0.7406 × 10^(-3) × 24

C = $0.0032 per day

Answer: As to the more sophisticated way of detecting "smoke" from an object a human may use in hotel rooms, this sensor called a Fresh Air Sensor does not just detect and, and but alerts the management about a smoking incident in a hotel room

Answer:

Printing press , lenses etc,

Explanation:

Almost all of the Renaissance innovations that influenced the world, printing press was one of them. The printing press is widely regarded as the Renaissance's most significant innovation. The press was invented by Johannes Gutenberg, and Germany's culture was altered forever.

Astrophysics, humanist psychology, the printed word, vernacular vocabulary of poetry, drawing, and sculpture practice are only a few of the Renaissance's main inventions.

Answer:

that not even a question

Answer:

The rate at which entropy is produced within the turbine is 22.762 kilojoules per kilogram-Kelvin.

Explanation:

By either the Principle of Mass Conservation and First and Second Laws of Thermodynamics, we model the steam turbine by the two equations described below:

Principle of Mass Conservation

[tex]\dot m_{in} - \dot m_{out} = 0[/tex] (1)

First Law of Thermodynamics

[tex]-\dot Q_{out} + \dot m \cdot \left[h_{in}-h_{out}+ \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) - w_{out} \right] = 0[/tex] (2)

Second Law of Thermodynamics

[tex]-\frac{\dot Q_{out}}{T_{out}} + \dot m\cdot (s_{in}-s_{out}) + \dot S_{gen} = 0[/tex] (3)

By dividing each each expression by [tex]\dot m[/tex], we have the following system of equations:

[tex]-q_{out} + h_{in}-h_{out} + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) - w_{out} = 0[/tex] (2b)

[tex]-\frac{q_{out}}{T_{out}} + s_{in}-s_{out} + s_{gen} = 0[/tex] (3b)

Where:

[tex]\dot Q_{out}[/tex] - Heat transfer rate between the turbine and its surroundings, in kilowatts.

[tex]q_{out}[/tex] - Specific heat transfer between the turbine and its surroundings, in kilojoules per kilogram.

[tex]T_{out}[/tex] - Outer surface temperature of the turbine, in Kelvin.

[tex]\dot m[/tex] - Mass flow rate through the turbine, in kilograms per second.

[tex]h_{in}[/tex], [tex]h_{out}[/tex] - Specific enthalpy of water at inlet and outlet, in kilojoules per kilogram.

[tex]v_{in}[/tex], [tex]v_{out}[/tex] - Speed of water at inlet and outlet, in meters per second.

[tex]w_{out}[/tex] - Specific work of the turbine, in kilojoules per kilogram.

[tex]s_{in}[/tex], [tex]s_{out}[/tex] - Specific entropy of water at inlet and outlet, in kilojoules per kilogram-Kelvin.

[tex]s_{gen}[/tex] - Specific generated entropy, in kilojoules per kilogram-Kelvin.

By property charts for steam, we get the following information:

Inlet

[tex]T = 400\,^{\circ}C[/tex], [tex]p = 3000\,kPa[/tex], [tex]h = 3231.7\,\frac{kJ}{kg}[/tex], [tex]s = 6.9235\,\frac{kJ}{kg\cdot K}[/tex]

Outlet

[tex]T = 100\,^{\circ}C[/tex], [tex]p = 101.42\,kPa[/tex], [tex]h = 2675.6\,\frac{kJ}{kg}[/tex], [tex]s = 7.3542\,\frac{kJ}{kg\cdot K}[/tex]

If we know that [tex]h_{in} = 3231.7\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 2675.6\,\frac{kJ}{kg}[/tex], [tex]v_{in} = 160\,\frac{m}{s}[/tex], [tex]v_{out} = 100\,\frac{m}{s}[/tex], [tex]w_{out} = 540\,\frac{kJ}{kg}[/tex], [tex]T_{out} = 350\,K[/tex], [tex]s_{in} = 6.9235\,\frac{kJ}{kg\cdot K}[/tex] and [tex]s_{out} = 7.3542\,\frac{kJ}{kg\cdot K}[/tex], then the rate at which entropy is produced withing the turbine is:

[tex]q_{out} = h_{in} - h_{out} + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2})-w_{out}[/tex]

[tex]q_{out} = 3231.7\,\frac{kJ}{kg} - 2675.6\,\frac{kJ}{kg} + \frac{1}{2}\cdot \left[\left(160\,\frac{m}{s} \right)^{2}-\left(100\,\frac{m}{s} \right)^{2}\right] - 540\,\frac{kJ}{kg}[/tex]

[tex]q_{out} = 7816.1\,\frac{kJ}{kg}[/tex]

[tex]s_{gen} = \frac{q_{out}}{T_{out}}+s_{out}-s_{in}[/tex]

[tex]s_{gen} = \frac{7816.1\,\frac{kJ}{kg} }{350\,K} + 7.3542\,\frac{kJ}{kg\cdot K} - 6.9235\,\frac{kJ}{kg\cdot K}[/tex]

[tex]s_{gen} = 22.762\,\frac{kJ}{kg\cdot K}[/tex]

The rate at which entropy is produced within the turbine is 22.762 kilojoules per kilogram-Kelvin.

Answer:

Null Hypothesis - True average compressive strength of  mixture of pulverized fuel ash and Portland cement is more than 1300 KN/m2.

Type I error (false positive) - this occur when the True average compressive strength of  mixture of pulverized fuel ash and Portland cement is not more than 1300 KN/m2 in actual scenario but the measurements shows an incorrect reading of True average compressive strength greater than 1300 KN/m2.

Type II error (false-negative)- his occur when the True average compressive strength of  mixture of pulverized fuel ash and Portland cement is  more than 1300 KN/m2 in actual scenario but the measurements show an incorrect  reading of True average compressive strength less than 1300 KN/m2.

Explanation:

The hypothesis for this piece of information is as follows -

Null Hypothesis - True average compressive strength of  mixture of pulverized fuel ash and Portland cement is more than 1300 KN/m2.

Alternate hypothesis - True average compressive strength of  mixture of pulverized fuel ash and Portland cement is less than or equal to 1300 KN/m2

Type I error (false positive) - this occur when the True average compressive strength of  mixture of pulverized fuel ash and Portland cement is not more than 1300 KN/m2 in actual scenario but the measurements shows an incorrect reading of True average compressive strength greater than 1300 KN/m2.

Type II error (false-negative)- his occur when the True average compressive strength of  mixture of pulverized fuel ash and Portland cement is  more than 1300 KN/m2 in actual scenario but the measurements show an incorrect  reading of True average compressive strength less than 1300 KN/m2.

Answer:

- 46.5171kW

Explanation:

FIrst, the value given:

P1 = 1.05 bar (Initial pressure)

P2 = 12 bar (final pressure)

Heat transfer, Q = - 3.5 kW (It is negative because the compressor losses heat to the surroundings)

Mgaseous nitrogen = Mair = 28.0134 Kg/mol (constant)

Universal gas constant, Ru = 8.3143 Kj/Kgmolk

Specific gas constant, R = 0.28699 Kj/KgK

Initial temperature, T1 = 300 K

Final temperature, T2 = 400 K

Finding the volume:

P1V1 = RT1

V1 = RT1 ÷ P1

= (0.28699 Kj/KgK X 300k) ÷ 105

Note convert bar to Kj/Nm by multiply it by 100

V1 =  0.81997 m3/Kg

To get the mass flow rate:

m = volumetric flow rate / V1

= (21 m3/min x 1/60seconds) ÷ 0.81997 m3/Kg

= 0.4268Kg/s

Using tables for the enthalpy,

hT1 = 300.19 KJ/Kg

hT2 = 400.98 KJ/Kg

The enthalpy change = hT2 - hT1

= 100.79 KJ/Kg

Power, P = Q - (m X enthalpy change)

= - 3.5 - (0.4268 X 100.79)

= - 46.5171kW

Answer:

[tex]D=0.1160m[/tex]

Explanation:

From the question we are told that:

Output Power [tex]P=1.2kw[/tex]

Density [tex]\rho=1.29kg/m^3[/tex]

Wind speed [tex]V=17.5mph=>7.8m/s[/tex]

Efficiency [tex]\gamma=60\%=>0.60[/tex]

Let Betz Limit

  [tex]C_p=\frac{16}{27}[/tex]

Generally the equation for Turbine Efficiency is mathematically given by

  [tex]\gamma=\frac{P}{P'}[/tex]

Where

P'=input power

 [tex]P' = In\ power[/tex]

 [tex]P'=1/2*C_p*\rho u^3*A[/tex]

 [tex]P'=1/2*C_p*\rho u^3*\frac{\pi}{4}*D^2[/tex]

Therefore the blade diameter for a two-blade propeller type rotor is

 [tex]\gamma=\frac{P*2*27*4}{16*\rho u^3*\pi*D^2}[/tex]

 [tex]D^2=\frac{P*2*27*4}{16*\rho u^3*\pi*\gamma}[/tex]

 [tex]D^2=\frac{1.2*2*27*4}{16*1.29*7.91^3*\pi*0.60}[/tex]

 [tex]D^2=0.0135[/tex]

 [tex]D=\sqrt{0.0135}[/tex]

 [tex]D=0.1160m[/tex]

Answer:

0.1160

Explanation: give them brainliest they deserve it

Answer:

International Organization For Standardization

Explanation:

This is the body within Geneva, Switzerland that promotes worldwide industrial and commercial standards

Answer:

Option A is correct

Explanation:

As we know  

Inductive Susceptance = ½(pi)*f*L

Or  Inductive Susceptance is inversely  proportional to the frequency

Likewise conductive Susceptance = 2 (pi)*f*C

Conductive Susceptance is directly proportional to the frequency

When the frequency will reach the value zero, then the Inductive Susceptance will become infinite

Hence, inductor will dominate in determining the equivalent impedance of this parallel combination

Option A