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3 years ago

10

Where is the sun located in the galaxy?

2 answers:

Anon25 [30]3 years ago

8 0

I believe it is with an outer arm

GarryVolchara [31]3 years ago

6 0

I think it's 3. within an outer arm

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What is one positive outcome of the recent<br>events?

fenix001 [56]

Answer:

Analyze the positive and negative consequences of catastrophic events of the last 40-50 years and the individuals that had an impact on these events. Meaning.

Explanation:

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2 years ago

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A power plant produces 1000 MW to supply a city 40 km away. Current flows from the power plant on a single wire of resistance 0.

nikitadnepr [17]

Answer:

Current = 8696 A

Fraction of power lost = $\dfrac{80}{529}$ = 0.151

Explanation:

Electric power is given by

$P=IV$

where I is the current and V is the voltage.

$I=\dfrac{P}{V}$

Using values from the question,

$I=\dfrac{1000\times10^6 \text{ W}}{115\times10^3\text{ V}} = 8696 \text{ A}$

The power loss is given by

$P_\text{loss} = I^2R$

where R is the resistance of the wire. From the question, the wire has a resistance of $0.050\Omega$ per km. Since resistance is proportional to length, the resistance of the wire is

$R = 0.050\times40 = 2\Omega$

Hence,

$P_\text{loss} = \left(\dfrac{200000}{23}\right)^2\times2$

The fraction lost = $\dfrac{P_\text{loss}}{P}=\left(\dfrac{200000}{23}\right)^2\times2\div (1000\times10^6)=\dfrac{80}{529}=0.151$

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2 years ago

Boron (B) has an atomic number of 5 and an atomic mass of 11. Boron has _____.

Neko [114]

Answer:

the answer is 5 electrons

Explanation:

because its the same name as the amount of protons

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3 years ago

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Acetone is a flammable solvent used in the making of plastics. The density is 790kg/m3. to convert 790kg/m3 to g/cm3

Artist 52 [7]

Answer: 0.790 g/cm3

Explanation:

The density of acetone is 790 Kg/m3.

To convert from Kg to g we multiply by 1000 (1 Kg = 1000 g)

To convert from m3 to cm3 we multiply by 10∧6

So, The density of acetone in (g/cm3) = (790 x 1000) / (10∧6) = 0.79 g/cm3

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2 years ago

Two light bulbs have resistances of 400 Ω and 800 ΩThe two light bulbs are connected in series across a 120- V line. Find the cu

Natasha2012 [34]

1) Current in each bulb: 0.1 A

The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:

$R_{eq}=R_1 + R_2 = 400 \Omega + 800 \Omega=1200 \Omega$

And so, the current through the circuit is (using Ohm's law):

$I=\frac{V}{R_{eq}}=\frac{120 V}{1200 \Omega}=0.1 A$

And since the two bulbs are connected in series, the current through each bulb is the same.

2) 4 W and 8 W

The power dissipated by each bulb is given by the formula:

$P=I^2 R$

where I is the current and R is the resistance.

For the first bulb:

$P_1 = (0.1 A)^2 (400 \Omega)=4 W$

For the second bulb:

$P_1 = (0.1 A)^2 (800 \Omega)=8 W$

3) 12 W

The total power dissipated in both bulbs is simply the sum of the power dissipated by each bulb, so:

$P_{tot} = P_1 + P_2 = 4 W + 8 W=12 W$

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2 years ago