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A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 35 m in 3.5 min, starting and ending at rest. The elevator's counterweight has a mass of only 940 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable

Answers

Answer 1

Answer:

425.1 W

Explanation:

We are given;

Counter mass of elevator; m_c = 940 kg

Cab mass of elevator; m_d = 1200 kg

Distance from rest upwards; d = 35 m

Time to cover distance; t = 3.5 min

Now, this elevator will have 3 forces acting on it namely;

Force due to the counter weight of the elevator; F_c

Force due to the cab weight on the elevator; F_d

Force exerted by the motor; F_m

Now, from Newton's 2nd law of motion,

The force exerted by the motor on the elevator can be given by the relationship;

F_m = F_d - F_c

Now,

F_d = m_d × g

F_d = 1200 × 9.81

F_d = 11772 N

F_c = m_c × g

F_c = 940 × 9.81

F_c = 9221.4 N

Thus;

F_m = 11772 - 9221.4

F_m = 2550.6 N

Now, the average power required of the force the motor exerts on the cab via the cable is given by;

P_m = F_m × v

Where v is the velocity of the elevator.

The velocity is calculated from;

v = distance/time

v = 35/3.5

v = 10 m/min

Converting to m/s gives;

v = 10/60 m/s = 1/6 m/s

Thus;

P_m = 2550.6 × 1/6

P_m = 425.1 W

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Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. For the water, determine the heat transfer, in kJ per kg of water. Kinetic and potential energy effects can be ignored.

Answers

Explanation:

Using table A-3, we will obtain the properties of saturated water as follows.

Hence, pressure is given as p = 4 bar.

$u_(1) = u_(g)$ = 2553.6 kJ/kg

$v_(1) = v_(g) = 0.4625 m^(3)/kg$

At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.

Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So, $v_(2) = v_(g) = 0.4625 m^(3)/kg$

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at $v_(2) = v_(g) = 0.4625 m^(3)/kg$ and temperature $T_(2) = 360^(o)C$ so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.

Now, using interpolation we will find the internal energy as follows.

$u_(2) = u_{\text{at 5 bar, 400^(o)C}} + (\frac{v_(2) - v_{\text{at 5 bar, 400^(o)C}}}{v_{\text{at 7 bar, 400^(o)C - v_(at 5 bar, 400^(o)C)}}})(u_{at 7 bar, 400^(o)C - u_(at 5 bar, 400^(o)C)})$

$u_(2) = 2963.2 + ((0.4625 - 0.6173)/(0.4397 - 0.6173))(2960.9 - 2963.2)$

= 2963.2 - 2.005

= 2961.195 kJ/kg

Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$ ......... (1)

Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.

$\Delta K.E = \Delta P.E$ = 0

Now, equation will be as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$

Q - 0 = $\Delta U + 0 + 0$

Q = $\Delta U$

Now, we will obtain the heat transfer per unit mass as follows.

$(Q)/(m) = \Delta u$

$(Q)/(m) = u_(2) - u_(1)$

= (2961.195 - 2553.6)

= 407.595 kJ/kg

Thus, we can conclude that the heat transfer is 407.595 kJ/kg.

Final answer:

The heat transfer is 227.4 kJ per kg of water.

Explanation:

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. To determine the heat transfer in kJ per kg of water, we need to calculate the heat absorbed by the water as it reaches 360°C.

Using the specific heat capacity of water (4,186 J/kg°C) and the change in temperature (360°C - 100°C), we can calculate the heat transfer:

Qw = mw * cw * AT = (1 kg) * (4186 J/kg°C) * (360°C - 100°C) = 227,440 J = 227.4 kJ

Therefore, the heat transfer is 227.4 kJ per kg of water.

Heat transfer is the process by which thermal energy moves from one object or substance to another due to a difference in temperature. This fundamental phenomenon occurs through three main mechanisms: conduction, convection, and radiation. Conduction involves the direct transfer of heat through a material, such as metal. Convection is the transfer of heat through the movement of fluids (liquids or gases). Radiation is the emission of electromagnetic waves carrying heat energy. Understanding heat transfer is essential in various fields, including physics, engineering, and environmental science, as it governs temperature regulation, climate dynamics, and the functioning of countless technological devices.

Learn more about Heat transfer here:

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An 800 kHz radio signal is detected at a point 2.1 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 800 mV/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. The intensity of the radio signal at that point is closest to

Answers

Answer:

$I=8.48* 10^(-4)\ W/m^2$

Explanation:

Given that,

Frequency of the radio signal, $f=800\ kHz=8* 10^5\ Hz$

It is detected at a pint 2.1 km from the transmitter tower, x = 2.1 km

The amplitude of the electric field is, E = 800 mV/m

Let I is the intensity of the radio signal at that point. Mathematically, it is given by :

$I=(E^2_(rms))/(c\mu_o)$

$E_(rms)$ is the rms value of electric field, $E_(rms)=(E)/(√(2) )$

$I=(E^2)/(2c\mu_o)$

$I=((800* 10^(-3))^2)/(2* 3* 10^8* 4\pi * 10^(-7))$

$I=8.48* 10^(-4)\ W/m^2$

So, the intensity of the radio signal at that point is $8.48* 10^(-4)\ W/m^2$ . Hence, this is the required solution.

An eighteen gauge copper wire has a nominal diameter of 1.02mm. This wire carries a constant current of 1.67A to a 200w lamp. The density of free electrons is 8.5 x 1028 electrons per cubic metre. Find the magnitude of:i. The current density ii. The drift velocity

Answers

Answer:

The current density is $J = 2.04 * 10^(6) A /m^2$

The drift velocity is $v_d = 1.5 * 10^(-4) m/s$

Explanation:

From the question we are told that

The nominal diameter of the wire is $d = 1.02 mm= (1.02)/(1000) = 0.00102 \ m$

The current carried by the wire is $I = 1.67 A$

The power rating of the lamp is $P = 200 W$

The density of electron is $n = 8.5 * 10^(28) \ e/m^3$

The current density is mathematically represented as

$J = (I)/(A)$

Where A is the area which is mathematically evaluated as

$A = \pi (d^2)/(4)$

Substituting values

$A = 3.142 * ((1.02 * 10^(-3))^2 )/(4)$

$A = 8.0*10^(-4)m^2$

$J = (1.67)/(8.0*10^(-4))$

$J = 2.04 * 10^(6) A /m^2$

The drift velocity is mathematically represented as

$v_d = (J)/(ne)$

Where e is the charge on one electron which has a value $e = 1.602 *10^(-19) C$

$v_d =(2.04 * 10^6 )/(8.5 *10^(28) * 1.6 * 10^(-19))$

$v_d = 1.5 * 10^(-4) m/s$

To remove 800j of heat the compressor in the fridge does 500j of work. how much heat is released into the room?

Answers

Answer:

Heat released into the room = 1300 J

Explanation:

CONCEPT:

According to second law of thermodynamics , heat cannot flow from a lower temperature to a higher temperature.But the refrigerator transfers heat from lower to higher temperature .For this , we have to do work on the refrigerator.

This work is used to transfer heat from lower to higher temperature.

Heat removed by fridge = 800 J

Work done = 500 J

Heat released into the room = Heat removed + work done

Heat released into the room = 800 +500

Heat released into the room = 1300 J

A baseball player throws a ball into the stands at 15.0 m/s and at an angle 45.0° above the horizontal. On its way down, the ball is caught by a spectator 4.10 m above the point where the ball was thrown. How much time did it take for the ball to reach the fan in the stands?

Answers

Answer:

Time = 1.61 seconds

Explanation:

Using the equation displacement of a trajectory motion in the y plane

Y = u t sin ů - ½gt²....equation 1 where

Y= vertical displacement =4.1

U = initial velocity = 15m/s

g = acc. Due to gravity = 10m/s

Ů = angle of trajectory = 45

t = time to reach fan on its way down

Sub into equ 1

4.1 = 15t sinů - ½ * 10t²

4.1 = 10.61t - 5t²

Solve using quadratic formula

t =[-B±( -B² -4AC)^½]/2A....equation 2

Where A = 5, B=10.61, C =4.1

Substitute A,B,C into equ2

t = (10.61±5.53)/10

t = 0.508seconds or 1.61seconds

Since it is on its way down t= 1.61 seconds

A barbell spins around a pivot at its center at A. The barbell consists of two small balls, each with mass 450 grams (0.45 kg), at the ends of a very low mass rod of length d = 20 cm (0.2 m; the radius of rotation is 0.1 m). The barbell spins clockwise with angular speed 120 radians/s.What is the speed of ball 1?