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KATRIN_1 [288]
3 years ago
8

In general, what happens to temperature as depth below earths surface increases

Physics
1 answer:
lesantik [10]3 years ago
8 0
Temperature increases as depth increases.
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taurus [48]
The answer is south
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2 years ago
An airplane has an average speed of 740 mph
crimeas [40]

Answer:

2590

Explanation:

7 0
3 years ago
Consider a parachutist that has reached terminal velocity. which of the following is true? A.) The acceleration of the parachuti
trasher [3.6K]
The correct answer is:
<span>B.) At terminal velocity there is no net force 

In fact, when the parachutist reaches the terminal velocity, his velocity does not change any more. It means that the acceleration acting on the parachutist is zero, and for Newton's second law, this means the net force acting on him is zero:
</span>\sum F = ma = 0
<span>because the acceleration is zero: a=0. 
This also means that the two relevant forces acting on the parachutist (gravity, downward, and air resistance, upward) are balanced to produce a net force equal to zero.</span>
8 0
3 years ago
Explain how sociology will contribute to the understanding of daily life and what is happening within it?
Helen [10]

Answer:

Explain how sociology will contribute to the understanding of daily life and what is happening within it?

Sociology is the study of society of human being, how they develop and those things put in place for such development

Explanation:

Sociology helps to develop culture, how people interact with one another and this boost the relationship and development as far as human society is concerned.

8 0
3 years ago
A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci
MA_775_DIABLO [31]

Answer:

v_B=3.78\times 10^5\ m/s

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is 6.63\times 10^{-27}\ kg

Speed of nucleus at A is v_A=6.2\times 10^5\ m/s

Potential at point A, V_A=1.5\times 10^3\ V

Potential at point B, V_B=4\times 10^3\ V

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s

So, the speed at point B is 3.78\times 10^5\ m/s.

7 0
3 years ago
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