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3 years ago

7

_____ have played a major role in altering Earth's atmospheric composition over time.A. TsunamisB. VolcanoesC. HurricanesD. Eart

hquakes

2 answers:

Lemur [1.5K]3 years ago

6 0

I think the correct answer is D.earthquakes.

sammy [17]3 years ago

4 0

I believe the answer is valcanos because they release sulfur in to the air and other chemicals so it willl affect the anispher

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The astronomer who first worked out the mathematical description of black hole event horizons was?

vovangra [49]

Karl Schwarzschild devised the first general relativity model that would adequately describe a black hole in 1916.

What is Black Hole?

A black hole is an area of spacetime with such intense gravitational pull that nothing can escape from it, not even light or other electromagnetic waves. According to general relativity theory, a compact enough mass can bend spacetime into a black hole. The event horizon is the line beyond which there is no escape.

Black holes were once thought to be a mathematical curiosity, but theoretical research in the 1960s revealed that they were actually a general prediction of general relativity.

To know more about Black Hole refer:

brainly.com/question/7866362

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4 0

2 years ago

Twenty grams of a solid at 70°C is place in 100 grams of a fluid at 20°C. Thermal equilibrium is reached at 30°C.

zaharov [31]

Answer:

c. is more than that of the fluid.

Explanation:

This problem is based on the conservation of energy and the concept of thermal equilibrium

$heat= m s \Delta T
$

m= mass

s= specific heat

\DeltaT=change in temperature

let s1= specific heat of solid and s2= specific heat of liquid

then

Heat lost by solid= $20(s_1)(70-30)=800s_1
$

Heat gained by fluid= $100(s_2)(30-20)=1000s_2
$

Now heat gained = heat lost

therefore,

1000 S_2=800 S_1

S_1=1.25 S_2

so the specific heat of solid is more than that of the fluid.

8 0

3 years ago

Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

Brrunno [24]

1) $+2.19\mu C$

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$ (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

8 0

3 years ago

Refraction can occur only when light is

mr_godi [17]

Answer:

We have learned that refraction occurs as light passes across the boundary between two media. Refraction is merely one of several possible boundary behaviors by which a light wave could behave when it encounters a new medium or an obstacle in its path.

3 0

2 years ago

What force would be required to accelerate a 1,100kg car to 0.5 m/s2

OlgaM077 [116]

Answer:

the force required to accelerate a 1,100kg car is 550N

5 0

3 years ago