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3 years ago

Resistance in a circuit that has 35 mA of current and 5.6 V applied equals.

Engineering

1 answer:

natka813 [3]3 years ago

5 0

Answer:

R = 160 Ω

Explanation:

For this problem, we will simply apply Ohm's law to find the resistance in the circuit. Ohm's law is as follows:

Voltage = Current x Resistance

We can solve the law for resistance:

Resistance = Voltage / Current

And now we just plug in our values:

Resistance = 5.6V / 35mA

Resistance = 5.6V / 0.035 A

Resistance = 160 Ω

Hence the resistance of the circuit is 160 Ω.

Cheers

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3–102 One of the common procedures in fitness programs is to determine the fat-to-muscle ratio of the body. This is based on the

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Answer:

x_fat = [ 0.5*(Wsa + Wsw) - p_muscle*V ] / V*( p_fat - p_muscle )

Explanation:

Given:

- The total volume of body = V

- The average density of the body = p_avg

- The density of muscle = p_muscle

- The density of fat = p_fat

Find:

Obtain a relation for the volume fraction of body fat x_fat

Solution:

- The volume of the fat is given by:

V_fat = x_fat*V

- The volume of the muscle is given by:

V_muscle = V - V_fat

= V - x_fat*V

=V*( 1 - x_fat )

- We will use the conservation of mass for the body related as:

mass_fat + mass_muscle = Total average mass

p_fat*V_fat + p_muscle*V_muscle = p_avg*V

p_fat*x_fat*V + p_muscle*V*( 1 - x_fat ) = p_avg*V

p_fat*x_fat + p_muscle*( 1 - x_fat ) = p_avg

- To determine p_1 we weigh the body in air:

Weight reading (Wsa) = m = p_1*V

p_1 = Wsa / V*g

- To determine p_2 we weigh the body in water:

Weight reading (Wsw) = m - p_w*V= p_1*V - p_w*V

Weight reading (Wsw) = V*(p_1 - p_w) = V*(p_2)

Where, p_2 = p_1 - p_water

p_2 = Wsw / V

- The average density p_avg:

p_avg = 0.5*(p_1 + p_2)

p_avg = 0.5*(Wsa / V + Wsw / V)

p_avg = 0.5*(Wsa + Wsw) / V

- Plug in the mass equation:

p_fat*x_fat + p_muscle*( 1 - x_fat ) = 0.5*(Wsa + Wsw) / V

x_fat*( p_fat - p_muscle ) = 0.5*(Wsa + Wsw) / V - p_muscle

x_fat = [ 0.5*(Wsa + Wsw) - p_muscle*V ] / V*( p_fat - p_muscle )

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3 years ago

Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost m

Ilia_Sergeevich [38]

Answer:

ordinary bulb total cost is $39.54

fluorescent bulb total cost is $13.05

amount save = 39.54 - 13.05 = $26.49

resistance = 626.1 ohm

Explanation:

in the 1st part

bulb on time = 3 year = 4380 hours

life of bulb = 750 h

so number of bulb required = $\frac{4380}{750}$

number of bulb required = 6

cost of 6 bulb is = 6 × 0.75 = $4.5

cost of operation is = 100 × 4380 × $\frac{0.08}{1000}$

cost of operation = $35.04

so total cost will be = $4.5 + $35.04 = $39.54

and

when compare with florescent bulb

time = 3 year = 4380 h

life of bulb = 10000 h

so number of bulb required = $\frac{4380}{10000}$

number of bulb required = 0.43 = 1

cost of 6 bulb is = 1 × 5 = $5

cost of operation is = 23 × 4380 × $\frac{0.08}{1000}$

cost of operation = $8.05

so total cost will be = $5 + $8.05 = $13.05

in part 2nd

total amount save while compare bulb is

amount save = 39.54 - 13.05 = $26.49

and in part 3rd

resistance of bulb is

resistance = $\frac{v^2}{P}$

resistance = $\frac{120^2}{23}$

resistance = 626.1 ohm

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Answer:

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Explanation:

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Answer:

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