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3 years ago

Which point refers to the epicenter of an earthquake?

Physics

2 answers:

alina1380 [7]3 years ago

6 0

Answer:

The area of the fault where the sudden rupture takes place is called the focus or hypocenter of the earthquake. The point on the Earth's surface directly above the focus is called the epicenter of the earthquake

Explanation:

Aleks04 [339]3 years ago

4 0

The epicenter is the point on the earth's surface vertically above the hypocenter. The answer is B

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Why is friction important for gymnast working on parallel bars

igomit [66]

Hey there,
Friction is important to increase the friction between the bars and the gymnast to prevent the gymnast from falling and hurting themselves.

Hope this helps :))

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3 years ago

A satellite in a circular orbit of radius R around planet X has an orbital period T. If Planet X had one-fourth as much mass, th

Iteru [2.4K]

<h2>Answer: 2T</h2>

According to the Third Kepler’s Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $R$ of its orbit.

This Law is originally expressed as follows (in the case of planet X and assuming we have a circular orbit):

$T^{2}=\frac{4\pi^{2}}{GM}R^{3}$ (1)

Where:

$G$ is the Gravitational Constant

$M=1.9(10)^{27}kg$ is the mass of planet X

$R$ is the radius of the orbit of the satellite around planet X

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=2\pi\sqrt{\frac{R^{3}}{GM}}$ (2)

Now, we are asked to find the period when tha mass of the planet is $\frac{1}{4}M$ . In order to do this, we have to rewrite equation (2) with this new value:

$T=2\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (3)

Solving:

$T=4\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (4)

On the other hand, if we multiply both sides of equation (2) by 2, we have:

$2T=4\pi\sqrt{\frac{R^{3}}{GM}}$ (5)

As we can see, (5) is equal to (4). This means the orbital period is twice the orignal period.

Hence, the answer is:

If Planet X had one-fourth as much mass, the orbital period of this satellite in an orbit of the same radius would be 2T.

3 0

3 years ago

Why does a solid have a definite shape and volume

Lady_Fox [76]

A solid has a definite shape and volume, since its particles are held closely together leaving no room.

4 0

3 years ago

Read 2 more answers

A river 1.00 mile wide flows with a constant speed of 1.00 mph. A man can row a boat at 2.00 mph. He crosses the river in a dire

gladu [14]

To solve this problem we will apply the geometric concepts of displacement according to the description given. Taking into account that there is an initial displacement towards the North and then towards the west, therefore the speed would be:

$V_T^2=v_N^2-v_W^2$

$V_T = \sqrt{v_N^2-v_W^2}$

Travel north 2mph and west to 1mph, then,

$V_T = \sqrt{2^2-1^2}$

$V_T = \sqrt{3}$

The route is done exactly the same to the south and east, so make this route twice, from the definition of speed we have to

$v= \frac{\Delta x}{t}$

$t = \frac{\Delta x}{v}$

$t = \frac{2*(1mile)}{\sqrt{3}mph}$

$t = 1.15hour$

Therefore the total travel time for the man is 1.15hour.

3 0

3 years ago

Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could als

lesantik [10]

Answer:

Explanation:

By convention, the electric field lines (which are tangent to the direction of the electric field at a given point) always begin at positive charges, and finish at negative charges.

This is a consequence of the convention that states that the electric field has the direction of the trajectory of a positive test charge when released from rest in an electric field.

(As the positive charge would move away from positive charges and would be attracted by negative ones).

So, the combination of answers that is true is b) (positive, negative, positive).

3 0

3 years ago