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Andrei [34K]
3 years ago
13

What kind of waves are present during an earthquake? ...

Physics
1 answer:
Brrunno [24]3 years ago
6 0

Answer:

1. The two main types of waves are body waves and surface waves. Body waves can travel through the earth's inner layers, but surface waves can only move along the surface of the planet like ripples on water. Earthquakes radiate seismic energy as both body and surface waves.

2. potential energy

3. Newton's second law of motion is F = ma, or force is equal to mass times acceleration.

4. Refraction is the bending of light

5. Density uses the formula p=m/V, or density (p) is equal to mass (m) divided by volume (V). Density is defined as mass per unit volume.

Explanation:

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A cylinder of compressed gas has a pressure of 488.2 kPa. The next day the cylinder of gas has a
Luda [366]

Answer:

20 °C

Explanation:

Ideal gas law:

PV = nRT

Rearranging:

P / T = nR / V

Since n, R, and V are constant:

P₁ / T₁ = P₂ / T₂

488.2 kPa / T = 468 kPa / 281.15 K

T = 293.29 K

T = 20.1 °C

Rounded, the temperature was 20 °C.

6 0
3 years ago
A solenoid with 435 turns has a length of 7.50 cm and a cross-sectional area of 3.50 ✕ 10−9 m2. Find the solenoid's inductance a
OLga [1]

Answer:

Solenoid's inductance is 1.11 × 10^-8H

The average emf around the solenoid is 1.3 × 10^-5V

Explanation: Please see the attachments below

4 0
3 years ago
According to Bernoulli's equation, the pressure in a fluid will tend to decrease if its velocity increases. Assuming that a wind
Pie

Answer:

The pressure drop predicted by Bernoulli's equation for a wind speed of 5 m/s

= 16.125 Pa

Explanation:

The Bernoulli's equation is essentially a law of conservation of energy.

It describes the change in pressure in relation to the changes in kinetic (velocity changes) and potential (elevation changes) energies.

For this question, we assume that the elevation changes are negligible; so, the Bernoulli's equation is reduced to a pressure change term and a change in kinetic energy term.

We also assume that the initial velocity of wind is 0 m/s.

This calculation is presented in the attached images to this solution.

Using the initial conditions of 0.645 Pa pressure drop and a wind speed of 1 m/s, we first calculate the density of our fluid; air.

The density is obtained to be 1.29 kg/m³.

Then, the second part of the question requires us to calculate the pressure drop for a wind speed of 5 m/s.

We then use the same formula, plugging in all the parameters, to calculate the pressure drop to be 16.125 Pa.

Hope this Helps!!!

7 0
3 years ago
Give a specific example of a vector quantity. why this type of quantity would be used to describe this event ?
Anarel [89]

example force. because you can say" I applied  3 newtons downward to the floor".

force has magnitude and direcion

3 0
3 years ago
The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c
eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

X_f = X_i +\frac{1}{2}(V_i+V_f)t

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

X_f = \frac{1}{2} (V_i+V_f)t

Clearing for time,

t = \frac{2X_f}{(V_i+V_f)}

t = \frac{2*1.2}{24+0}

t = 0.1s

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

a= \frac{\Delta V}{t}

Then

F = m\frac{\Delta V}{t}

F = m\frac{V_f-V_i}{t}

F = m\frac{-V_i}{t}

F = \frac{(1600kg)(-24m/s)}{(0.1s)}

F = -384000N

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

V_f^2=V_i^2-2ax

0 = (24m/s)^2-2*a(1.2m)

a = \frac{(24m/s)^2}{1.2m}

a=480m/s^2

The gravity is equal to 0.8, then the acceleration is

a = 480*\frac{g}{9.8}

a = 53.3g

3 0
3 years ago
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