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NikAS [45]
3 years ago
7

Move the magnet at a relatively constant frequency back and forth through the coil. The voltage displayed is proportional to the

current flowing in the coil. What happens as you move the magnet through the coils with different number of loops?
Physics
1 answer:
Jet001 [13]3 years ago
3 0

Answer:

<em>The induced EMF, and hence the induced current produced will increase or decrease with the number of loops on the coil.</em>

Explanation:

According to Faraday' law of electromagnetic induction, the induced EMF increases with the speed with which we turn the coil, the surface area of the coil, the number of loops, and the strength of the magnetic field. From this, we can see that increasing the number of loops also increases the surface area involved. This means that if we move the magnet in this experiment through the coils with different number of loops, <em>the induced EMF, and hence the induced current, will increase or decrease with an increase or decrease  in the number of loops respectively. </em>

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A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done
Elena-2011 [213]

Answer:

The work done by gravity during the roll is 490.6 J

Explanation:

The work (W) is:

W = F*d

<em>Where</em>:

F: is the force

d: is the displacement = 20 m

The force is equal to the weight (W) in the x component:

F = W_{x} = mgsin(\theta)

<em>Where:</em>

m: is the mass of the bowling ball = 5 kg

g: is the gravity = 9.81 m/s²    

θ: is the degree angle to the horizontal = 30°        

F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N    

Now, we can find the work:

W = F*d = 24.53 N*20 m = 490.6 J      

Therefore, the work done by gravity during the roll is 490.6 J.

I hope it helps you!

6 0
3 years ago
Read the scenario below and answer the question that follows. Randall is hiring cooks for his restaurant. The first applicant is
Talja [164]
Randall has unconscious assumption that attractive people are more competent
4 0
3 years ago
Given the function f(x) = 8x3 − 3x2 − 5x 8, what part of the function indicates that the left and right ends point in opposite d
ale4655 [162]

Answer: The degree of the first term.

Explanation:

The function:

f(x) = 8x^3-3x^2-5x^8

The left and right ends would be indicated when x is changed to -x. When this is substituted, the change is indicated by the first term because only the degree of first term is odd.

Let the left hand side be donated by -x.

Then,

f(-x) =8(-x)^3-3(-x)^2-5(-x)^8\\ \Rightarrow f(-x)=-8x^3-3x^2-5x^8

Hence, the correct option is the degree of the first term indicates the left and right end points of the function.

8 0
3 years ago
Read 2 more answers
A,b, e are complete. Help on the others would be so appreciated!!
bixtya [17]

Answer:

serie Ceq=0.678 10⁻⁶ F  and the charge Q = 9.49 10⁻⁶ C

Explanation:

Let's calculate all capacity values

a) The equivalent capacitance of series capacitors

    1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3 + 1 / C4 + 1 / C5

    1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 1 / 6.2 + 1 / 6.2

    1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 2 / 6.2

    1 / Ceq = 0.666 + 0.3030 +0.1818 +0.3225

    1 / Ceq = 1,147

    Ceq = 0.678 10⁻⁶ F

b) Let's calculate the total system load

   Dv = Q / Ceq

   Q = DV Ceq

   Q = 14 0.678 10⁻⁶

   Q = 9.49 10⁻⁶ C

In a series system the load is constant in all capacitors, therefore, the load in capacitor 5.5 is Q = 9.49 10⁻⁶ C

c) The potential difference

   ΔV = Q / C5

   ΔV = 9.49 10⁻⁶ / 5.5 10⁻⁶

   ΔV = 1,725 ​​V

d) The energy stores is

    U = ½ C V²

    U = ½ 0.678 10-6 14²

    U = 66.4 10⁻⁶ J

e) Parallel system

   Ceq = C1 + C2 + C3 + C4 + C5

   Ceq = (1.5 +3.3 +5.5 +6.2 +6.2) 10⁻⁶

   Ceq = 22.7 10⁻⁶ F

f) In the parallel system the voltage is maintained

   Q5 = C5 V

   Q5 = 5.5 10⁻⁶ 14

   Q5 = 77 10⁻⁶ C

g) The voltage is constant V5 = 14 V

h) Energy stores

   U = ½ C V²

   U = ½ 22.7 10-6 14²

   U = 2.2 10⁻³ J

8 0
3 years ago
One of China's cement factories has become more energy efficient by ________. a. using the extra gas and heat from the kilns to
OleMash [197]

Answer:

a. using the extra gas and heat from the kilns to generate electricity

Explanation:

Wiley Online Library

Energy Science & EngineeringVolume 5, Issue 2

Research Article Open Access

The generation of power from a cement kiln waste gases: a case study of a plant in Kenya

Stanley Ngari Irungu Peter Muchiri Jean Bosco Byiringiro

First published: 01 April 2017

https://doi.org/10.1002/ese3.153

Citations: 1

No funding information provided.

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Abstract

The cement production process is energy intensive both in terms of the thermal energy (firing the kiln, drying and De carbonation) and electrical energy for driving the numerous drives within the process line. The average specific power consumption of the case study plant was 111 kWh/ton of cement with an average peak demand of 9.7 MW. The high cost of electric power at 0.14 USD/kWh results in very high cost of production that significantly lowers the company's profit margin and limits its competitive advantage. The generation of electrical power from waste heat recovery would reduce the electricity power bill through partially substituting the power procured from the national grid. This research evaluated the potential that the plant has for generating electrical power from the hot waste gases vented into the atmosphere and it was found that the plant has the potential to generate 3.4 MWh of electrical power. This results to a net potential to generate 2.89 MWh of electrical power after factoring in the auxiliary power consumption by Waste heat recovery plant system at 15%. This ultimately gave a reduction of 33% in the electricity power bill of the case study plant. The paper recommends the installation of a steam rankine cycle for the power generating plant. In this work the authors designed the steam boilers for the waste heat recovery plant for conversion of thermal energy to electrical energy, selected a commercial steam turbine and evaluated its economic feasibility and established that the designed plant would have a simple payback period of 2.7 years.

Introduction

The cement manufacturing process is an energy intensive industry, both in terms of thermal and electrical energy. The cost of energy keeps on fluctuating and this negatively impact on the manufacturing cost and eventually lowers the competitiveness and profitability of the cement industry. The energy costs in a cement industry account for about 26% of the total manufacturing cost of cement which is in the form of electrical energy accounting for 25% of the input energy and 75% is thermal energy 1. Furthermore, the sources of thermal energy utilized in the cement industry are mostly nonrenewable and this necessitates deep consideration of energy conservation to guarantee sustainability.

The case study plant suffers financial loss as a result of higher per unit cost of power from the grid and the poor reliability of the supply. The poor reliability of supply negatively affects the kiln operations (the heart of operations) as a result of the sensitivity of the process to power quality resulting in high set up costs. This significantly raises the cost of production for the case study plant and eventually results in the loss of her competitive advantage.

The generation of Power from the cement kiln Waste Heat gases is an energy saving opportunity and it entails the recovery of the heat energy contained in the waste gases that are emitted into the atmosphere from the cement kiln. According to 2, the generation of Power from kiln Waste Heat Recovery is about conversion of the waste heat from the clinkering process into useful electrical energy. Cogeneration of power is achieved by utilizing this waste heat streams from the preheater and the cooler, passing the waste gases through boilers, which in turn generate steam which is used to turn/run turbines to generate electricity

7 0
3 years ago
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