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kiruha [24]
3 years ago
14

What is the main evidence for the structure of the Earth that occur from the generation of an earthquake?

Physics
2 answers:
Alik [6]3 years ago
4 0

Answer;

-Waves

Explanation;

-The main evidence for the structure of the Earth comes from studying waves generated during an earthquake. When stress builds up and rocks at depth break, then elastic waves, just like sound waves are sent from the point of rupture. These waves speed out and move through the earth.

-When seismometers are put on the surface of the Earth, these waves can be measured. In particular, if we have enough seismic stations and the quake is big enough, then we can take all the travel time information and invert it. That means we can obtain a graph that shows the density of rock as a function of depth.

-Once we know the density we can estimate what the actual materials are. We find that the earth has a small inner core made out of iron surrounded by a larger outer liquid iron core and that is surrounded by a rocky mantle and rocky crust.

xeze [42]3 years ago
3 0
I believe that its waves

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Explain about ohm's law.​
belka [17]

Answer:

Statement:

The electric current passing through a conductor is directly proportional to the potential difference across its ends provided temperature and other physical conditions remain constant.

Explanation:

Current is directly proportional to voltage loss through a resistor. That is, if the current doubles, then so does the voltage. To make a current flow through a resistance there must be a voltage across that resistance. Ohm's Law shows the relationship between the voltage (V), current (I) and resistance (R).

V∝I or I∝V⇒V=IR.

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3 years ago
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What are the different types of topology?​
dolphi86 [110]

Answer:

common types of topologies, and we're going to break each of them down in the guide below.

Bus topology. As the simplest design, a bus topology requires nodes to be in a linear order. ...

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2 years ago
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An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 60 cm long and has a mass of 3.8 kg, with the center of
Serggg [28]

Answer:

(a) τ = 26.58 Nm

(b) τ = 18.79 Nm

Explanation:

(a)

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

d₁ = perpendicular distance between ball and shoulder = 60 cm = 0.6 m

τ₁ = (29.4 N)(0.6 m)

τ₁ = 17.64 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of 60 cm = (0.4)(60 cm) = 24 cm = 0.24 m

τ₂ = (37.24 N)(0.24 m)

τ₂ = 8.94 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 17.64 Nm + 8.94 Nm

<u>τ = 26.58 Nm</u>

<u></u>

(b)

Now, the arm is at 45° below horizontal line.

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

42.42 cm = 0.4242 m

τ₁ = (29.4 N)(0.4242 m)

τ₁ = 12.47 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of (60 cm)(Cos 45°) = (0.4)(42.42 cm) = 16.96 cm = 0.1696 m

τ₂ = (37.24 N)(0.1696 m)

τ₂ = 6.32 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 12.47 Nm + 6.32 Nm

<u>τ = 18.79 Nm</u>

3 0
2 years ago
Estimate the moment of inertia of a bicycle wheel 66 cm in diameter. The rim and tire have a combined mass of 1.2 Kg
Fed [463]
<span>For a point mass the moment of inertia is just the mass times the square of perpendicular distance to the rotation axis, I = mr^2. That point mass relationship becomes the basis for all other moments of inertia since any object can be built up from a collection of point masses. So the I = (1.2 kg)(0.66m/2)^2 = 0.1307 kg m2</span>
3 0
3 years ago
A white dwarf star has a density of about 1.0 x 10^9 kg/m3. If the earth were to suddenly become as dense as a white dwarf star,
GalinKa [24]

Answer:

R = 98304.75 m = 98.3 km

Explanation:

The density of an object is given as the ratio between the mass of that object and the volume occupied by that object.

Density = Mass/Volume

Now, it is given that the density of Earth has become:

Density = 1 x 10⁹ kg/m³

Mass = Mass of Earth (Constant) = 5.97 x 10²⁴ kg

Volume = 4/3πR³ (Volume of Sphere)

R = Radius of Earth = ?

Therefore,

1 x 10⁹ kg/m³ = (5.97 x 10²⁴ kg)/[4/3πR³]

4/3πR³ = (5.97 x 10²⁴ kg)/(1 x 10⁹ kg/m³)

R³ = (3/4)(5.97 x 10¹⁵ m³)/π

R = ∛[0.95 x 10¹⁵ m³]

<u>R = 98304.75 m = 98.3 km</u>

6 0
2 years ago
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