Ask question

Ask question

All categories

2 years ago

9

The current in a circuit is tripled by connecting a 580 resistor in parallel with the resistance of the circuit. Determine the r

esistance of the circuit in the absence of the 580 resistor.

1 answer:

const2013 [10]2 years ago

6 0

Answer:

1160 ohm

Explanation:

We are given that

R'=580 ohm

Current=3 I

We have to find the resistance of the circuit.

Let R be the resistance of circuit.

In parallel

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

Using the formula

$\frac{1}{R_{eq}}=\frac{1}{580}+\frac{1}{R}=\frac{580+R}{580R}$

$R_{eq}=\frac{580R}{580+R}$

In parallel combination,Potential difference across each resistance remains same.

$V=IR$

Using the formula

$IR=3IR_{eq}$

$IR=3I\times \frac{580R}{580+R}$

$580+R=3\times 580=1740$

$R=1740-580=1160\Omega$

You might be interested in

Which of the following changes would make a heat engine waste more energy as heat

Komok [63]

A decrease in it's operating temperature would make a heat engine less efficient. This is because in order to operate, a heat engine needs to be hot and maintain that temperature.

4 0

3 years ago

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a

zepelin [54]

Answer:

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.

Explanation:

Given Values:

L = 50 cm = 0.5 m

H = 170 j/s

To find the diameter of the rod, we have to find the area of the rod using the following formula.

Here Tc = 100.0° C

k = 50.2

H = k × A × $\frac{[T_{H -}T_{C} ] }{L}$

Solving for A

A = $\frac{H * L }{k * [ T_{H}- T_{C} ] }$

A = $\frac{170 * 0.5}{50.2 * [ 350 - 100 ]}$

A = $\frac{85}{12550}$ = 6.77 × $10^{-3}$ m²

Now Area of cylinder is :

A = $\frac{\pi }{4}$ d²

solving for d:

d = $\sqrt{\frac{4 * 0.00677 }{\pi } }$

d = 9.28 cm

5 0

3 years ago

You decide to visit Santa Claus at the north pole to put in a good word about your splendid behavior throughout the year. While

Fynjy0 [20]

Answer:

Explanation:

The question here is that if sneezy hands from a similar rope while delivering presents at the earth's equator, what will be the tension in the rope be. Here is the solution:The tension on the rope when it is at pole, T= 455 NTo find, the tension, t= mgTo solve for mass, m= t/g. Substituting this we have, m=455/9.8. m=46.43 kgAssume that the downwards acceleration is, a= -46.43 m/s^2.T = mg + maT = (46.43 kg) ( 9.8 m/s^2) - (46.43 kg) (-46.43 m/s^2)T = 455.01 kg-m/s^2 - -2155.74 kg-m/s^2T = 2610.75 kg-m/s^2 = 2610.75 N

3 0

2 years ago

Materials that conduct heat and electricity well in the solid state result when

MA_775_DIABLO [31]

Materials that conduct heat and electricity well in the solid state result when metals bond with metals. <span>This type of bonding is called metallic bonding. Metallic bonding is when positive ions (metals) are in a 'sea of negative electrons'. The electrons are delocalised, which means they can move around easily and carry charge, and this enables it to conduct electricity, even in a solid state.</span>

8 0

3 years ago

A truck is traveling at 27 m/s down the interstate highway where you are changing a flat tire. frequency of 185 Hz.

Reil [10]

Answer:

(a) the observed frequency is 200 Hz

(b) the observed frequency is 188 Hz.

Explanation:

speed of the truck, Vs = 27 m/s

frequency of the truck as it approaches, Fs = 185 Hz

(a) Apply Doppler effect to determine the frequency you will hear.

As the truck approaches you, the observed frequency will be higher than the source frequency because of decrease in distance.

$F_s = F_o [\frac{V}{V_S + V} ]$

Where;

Fo is the observed frequency which is the frequency you will hear.

V is speed of sound in air

$F_s = F_o [\frac{V}{V_S + V} ]\\\\185 = F_o [\frac{340}{27 + 340} ]\\\\185 = F_o (0.926)\\\\F_o = \frac{185}{0.926}\\\\F_o = 199.78 \ Hz$

$F_o = 200 \ Hz$

(b) Apply the following formula for a moving observer and a moving source;

$F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )$

The observed frequency is negative since you are driving away from the truck and the source frequency is also negative since it is driving towards you.

$F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )\\\\F_o = 185[\frac{340-22}{340} ](\frac{340}{340-27} )\\\\F_o = 185(0.9353)(1.0863)\\\\F_o = 188 \ Hz$

5 0

3 years ago