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antiseptic1488 [7]
3 years ago
14

Someone help me asap!

Physics
2 answers:
gulaghasi [49]3 years ago
5 0
The answer is b .......
bogdanovich [222]3 years ago
4 0

Answer:the answer would be B, heavy rainfall causes a landslide on the side of the mountain.

Explanation:

Hydrosphere is all water or liquid elements in the earths atmosphere and geosphere is the solid parts, such as rock.

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I need help with 2 questions . Please help c:
dezoksy [38]
A) According to the nebular theory, the Solar System formed from a huge gaseous nebula which at a certain point was perturbated. Atoms and molecules started colliding, forming planetesimals (a sort of big rocks). The planetesimals were attracted to each other by gravity, forming bigger warm almost spherical objects called protoplanets, which at the end cooled down forming planets.
Therefore the correct answer is "all of the above".

b) The planets closer to the Sun were (and still are) subject to higher temperatures, due to their close distance to the Sun. In these conditions, rocky materials undergo condensation, while iced gaseous materials undergo vaporization. In the outer parts of the Solar System temperatures are too low to allow these transformations.
The correct answer is again "all of the above".
8 0
3 years ago
Which of the following accurately shows how to calculate the weight of a 20kg object
forsale [732]
The answer to your question is "20kgx9.8m/s" because weight is the force an object is exerting on another object, and the formula used to calculate force is <em>Force = Mass * Acceleration</em>.
4 0
3 years ago
Read 2 more answers
69PTS! Which choices are also major levels of classification? Choose all answers that are correct. A. Genus B. Species C. Group
Pachacha [2.7K]
A. Genus and B. Species are also major levels of classification.
5 0
3 years ago
Read 2 more answers
Which Coefficients balance the following equation? ____Pb(OH)2+_____HCI—&gt; _____ H2O+______PbCl2
lorasvet [3.4K]

the answer is A it is so easy question

3 0
3 years ago
What is the magnitude of the applied electric field inside an aluminum wire of radius 1.2 mm that carries a 3.0-a current? [ σal
galben [10]
Hello

1) First of all, since we know the radius of the wire (r=1.2~mm=0.0012~m), we can calculate its cross-sectional area
A=\pi r^2 = 3.14 \cdot (0.0012~m)^2=4.5\cdot10^{-6}~m^2

2)  Then, we can calculate the current density J inside the wire. Since we know the current, I=3~A, and the area calculated at the previous step, we have
J= \frac{I}{A}= \frac{3~A}{4.5\cdot10^{-6}~m^2} = 6.63\cdot10^5 ~A/m^2

3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity \sigma=3.6\cdot10^7~ \frac{A}{Vm} of the aluminium, the electric field is given by
E= \frac{J}{\sigma}= \frac{ 6.63\cdot10^5 ~A/m^2}{3.6\cdot10^7~ \frac{A}{Vm} } = 0.018~V/m

4 0
3 years ago
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