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Vesnalui [34]
3 years ago
8

Two AAA-size lithium batteries are connected in series in a flashlight. Each battery has 3.5 volt and 4- Amp-hour capacity. If t

he flashlight is connected to a light bulb with 10 Ohms resistance. How long does it take to drain the batteries?
Engineering
1 answer:
mafiozo [28]3 years ago
7 0

Answer:

t= 9.79 hr

Explanation:

Given that

V= 3.5 V

Capacity= 4 Amp-hour

We know that

V= IR

V= Voltage

I =Current

R=Resistance

V = I R

The total voltage on the  batteries will be  2 V

2 x 3.5 = I x 10

I= 0.7 A

We know that Power P

P = V I

P = 0.7 x 7

P =4.9 W

4 A.h and 12 volt power supply = 4 x 12 = 48 W.hr

So time of drain t

4.9 t = 48

t= 9.79 hr

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a. Determine R for a series RC high-pass filter with a cutoff frequency (fc) of 8 kHz. Use a 100 nF capacitor. b. Draw the schem
Readme [11.4K]

Answer:

a) 199.04 ohms

b) attached in image

c) -0.696dB

Explanation:

We are given:

Fc = 8Khz = 8000hz

C = 100nF = 100*10^-^9F

a)Using the formula:

F_c = \frac{1}{2pie*Rc}

8000= \frac{1}{2*3.14*R*100*10^-^9}

R =\frac{1}{2*3.14*100*10^-^9*8000}

R = 199.04 ohms

b) diagram is attached

c) H(w) = \frac{V_out(w)}{Vin(w)} = \frac{1}{1-j\frac{wc}{w}}

H(F) = \frac{1}{1-j\frac{fc}{f}}

At F = 20KHz and Fc= 8KHz we have:

H(F)= \frac{1}{1-j\frac{8}{20}} = \frac{1}{1-j(0.4)}

|H(F)|= \frac{1}{\sqrt{1^2+(0.4)^2}}

=0.923

|H(F)| in dB = 20log |H(F)|

=20log0.923

= -0.696dB

5 0
3 years ago
Given the inherent costs of regulation it is safe to say that there is always a negative economic impact associated with regulat
Alexus [3.1K]

the answer is true.                                     <u>                    </u>                                                              

6 0
2 years ago
A heat pump receives heat from a lake that has an average wintertime temperature of 6o C and supplies heat into a house having a
Dafna1 [17]

Answer:

a) \dot W = 1.062\,kW

Explanation:

a) Let consider that heat pump is reversible, so that the Coefficient of Performance is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

COP_{HP} = \frac{298.15\,K}{298.15\,K-279.15\,K}

COP_{HP} = 15.692

The minimum heat received by the house must be equal to the heat lost to keep the average temperature constant. Hence:

\dot Q_{H} = 60000\,\frac{kJ}{h}

The minimum power supplied to the heat pump is:

\dot W = \frac{\dot Q_{H}}{COP}

\dot W = \frac{\left(60000\,\frac{kJ}{h}  \right)\cdot \left(\frac{1\,h}{3600\,s}  \right)}{15.692}

\dot W = 1.062\,kW

5 0
3 years ago
What is the reason for the development of new construction of materials for human use? (Select all that apply.)
salantis [7]

Answer:

All 4 could be justified.

Explanation:

They all represent ultimate improvement.

7 0
2 years ago
Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7 m/s. Assume the width of the plate (into the paper)
mezya [45]

Complete Question

Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7m/s. Assume the width of the plate (into the paper) is 0.5 m. If the plate is at a co temperature of 100C,find:

The total heat transfer rate from the plate to the air

Answer:

q=1.7845

Explanation:

From the question we are told that:

Air Temperature T_1=40c

Length l=2m

Velocity v=7m/s

Width w=0.5

Constant temperature T_t= 100C

Generally the equation for Total heat Transfer is mathematically given by

 q=hA(T_s-T_\infty)

Where

h=Convective heat transfer coefficient

 h=29.9075w/m^2k

Therefore

 q=h(L*B)(T_s-T_\infty)

 q=29.9075*(2*0.5)(100+273-(40+273))

 q=1794.45w

 q=1.7845

5 0
3 years ago
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