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3 years ago

What does fissure and conduit have in common?

Physics

1 answer:

Flura [38]3 years ago

8 0

Answer:

although commons their deposits can dense

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Make the following conversion. 425 g = _____kg 0.425 0.0425 4,250 425,000

blsea [12.9K]

0.425
BECAUSE 1000g makes 1Kg
so 425g will make 0.425 option a

6 0

4 years ago

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Alan starts from his home and walks 1.3 km east to the library. He walks an additional 0.68 km east to a music store. From there

zepelin [54]

Answer: final Displacement = 0 km, total distance covered =7 km

Explanation:

Given the following :

From home to library = 1.3 km East

Library to music store = 0.68km East

Music store to friend's house = 1.1km North

Friend's house to grocery store = 0.42 km North

Displacement is the net change in distance traveled.

Eastward distance :

To = (1.3 + 0.68)km = 1.98km East

Fro = (0.68 + 1.3)km = 1.98 km East

Δ distance = (1.98 - 1.98) = 0

Northward direction:

To = (1.1 + 0.42)km = 1.52km north

Fro = (0.42 + 1.1)km = 1.52km North

Δ distance = (1.98 - 1.98) = 0

Hence final Displacement = 0

Total distance covered = 2 × (1.3 + 0.68 + 1.1 + 0.42) = 2 × 3.5

= 7km

3 0

4 years ago

Part A

Blababa [14]

Answer:

The wavelength decreases as the electron speeds up. The wavelength increases as the electron slows down.

Explanation:

5 0

3 years ago

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

3 0

3 years ago

What type of equilibrium maintains body position during sudden motion?

Juli2301 [7.4K]

the correct answer is dynamic

6 0

3 years ago

Read 2 more answers